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Class 10 Maths Part 1 Chapter 3 Arithmetic Progression Problem Set 3 Solutions | Maharashtra Board

Master Class 10 Maths with these comprehensive Textbook Exercise Solutions for the Maharashtra Board. This guide covers every question from Chapter 3 Arithmetic Progression Problem Set 3 (Maths Part 1). Whether you are preparing for your SSC board exams or revising core concepts, these step-by-step solutions, highlighted formulas, and expert exam tips will help you secure full marks. Let's dive into the solutions!

Question 1: Choose the correct alternative answer for each of the following sub questions.

i. The sequence -10, -6, -2, 2, ...

is an A.P. Reason $ d = -16 $
is an A.P. Reason $ d = 4 $
is an A.P. Reason $ d = -4 $
is not an A.P.

Answer: (B)

Note: The textbook misprinted the first term as 10. For the sequence to be an A.P. with the correct given option (B), the first term must be -10. $ d = -6 - (-10) = 4 $.

ii. First four terms of an A.P. are ..., whose first term is -2 and common difference is -2.

-2, 0, 2, 4
-2, 4, -8, 16
-2, -4, -6, -8
-2, -4, -8, -16

Answer: (C)

iii. What is the sum of the first 30 natural numbers?

Answer: (B)

Hint: $ S_{30} = \frac{30}{2}(1 + 30) = 15 \times 31 = 465 $

iv. For a given A.P., $ t_{7} = 4 $, $ d = -4 $, then $ a = $

Answer: (D)

Hint: $ t_7 = a + 6d \implies 4 = a + 6(-4) \implies a = 4 + 24 = 28 $

v. For a given A.P., $ a = 3.5 $, $ d = 0 $, $ n = 101 $, then $ t_{n} = $

0
3.5
103.5
104.5

Answer: (B)

vi. In an A.P., the first two terms are -3, 4, then the $ 21^{st} $ term is ....

-143
143
137
17

Answer: (C)

Hint: $ a = -3 $, $ d = 4 - (-3) = 7 $. $ t_{21} = -3 + 20(7) = -3 + 140 = 137 $

vii. If for any A.P. $ d = 5 $, then $ t_{18} - t_{13} = \dots $

Answer: (C)

Hint: $ t_{18} - t_{13} = (a + 17d) - (a + 12d) = 5d = 5 \times 5 = 25 $

viii. Sum of first five multiples of 3 is ...

Answer: (A)

ix. 15, 10, 5, ... In this A.P. sum of first 10 terms is...

-75
-125
75
125

Answer: (A)

x. In an A.P. $ 1^{st} $ term is 1 and the last term is 20. The sum of all terms is 399, then $ n = $

Answer: (B)

Hint: $ S_n = \frac{n}{2}(t_1 + t_n) \implies 399 = \frac{n}{2}(1 + 20) \implies n = \frac{798}{21} = 38 $

Question 2

Find the fourth term from the end in an A.P.: $ -11, -8, -5, \dots, 49 $

Solution:

Given: The A.P. is $ -11, -8, -5, \dots, 49 $
To Find: The $ 4^{th} $ term from the end.

Exam Tip: To find a term from the end easily, simply reverse the entire sequence. The new first term becomes the original last term, and the common difference reverses its sign!

Calculation:

Reversing the A.P., we get: $ 49, \dots, -5, -8, -11 $

Here, the new first term $ a = 49 $

The new common difference $ d = -11 - (-8) = -11 + 8 = -3 $

Formula:

\[ t_n = a + (n-1)d \]

Substitution:

$ t_4 = 49 + (4 - 1)(-3) $

$ t_4 = 49 + (3)(-3) $

$ t_4 = 49 - 9 $

$ t_4 = 40 $

Final Answer:

The fourth term from the end in the given A.P. is 40.

Question 3

In an A.P. the $ 10^{th} $ term is 46, sum of the $ 5^{th} $ and $ 7^{th} $ term is 52. Find the A.P.

Solution:

Given: $ t_{10} = 46 $ and $ t_5 + t_7 = 52 $
To Find: The sequence of the A.P.

Formula:

\[ t_n = a + (n-1)d \]

Calculation:

For an A.P., let $ a $ be the first term and $ d $ be the common difference.

$ t_{10} = a + (10 - 1)d $

$ 46 = a + 9d $ ... (i)

Also, $ t_5 + t_7 = 52 $

$ [a + (5 - 1)d] + [a + (7 - 1)d] = 52 $

$ a + 4d + a + 6d = 52 $

$ 2a + 10d = 52 $

Dividing throughout by 2:

$ a + 5d = 26 $ ... (ii)

Subtracting equation (ii) from (i), we get:

\[ (a + 9d) - (a + 5d) = 46 - 26 \] \[ 4d = 20 \implies d = 5 \]

Substituting $ d = 5 $ in equation (ii):

$ a + 5(5) = 26 $

$ a + 25 = 26 \implies a = 1 $

Now, let's find the terms of the A.P.:

$ t_1 = a = 1 $
$ t_2 = t_1 + d = 1 + 5 = 6 $
$ t_3 = t_2 + d = 6 + 5 = 11 $
$ t_4 = t_3 + d = 11 + 5 = 16 $

Final Answer:

The required A.P. is 1, 6, 11, 16, ...

Question 4

The A.P. in which $ 4^{th} $ term is -15 and $ 9^{th} $ term is -30. Find the sum of the first 10 numbers.

Solution:

Given: $ t_4 = -15 $ and $ t_9 = -30 $
To Find: $ S_{10} $ (Sum of the first 10 terms)

Key Formulas:
$ t_n = a + (n-1)d $
$ S_n = \frac{n}{2}[2a + (n-1)d] $

Calculation:

$ t_4 = a + 3d = -15 $ ... (i)

$ t_9 = a + 8d = -30 $ ... (ii)

Subtracting equation (i) from (ii):

\[ (a + 8d) - (a + 3d) = -30 - (-15) \] \[ 5d = -15 \implies d = -3 \]

Substituting $ d = -3 $ in equation (i):

$ a + 3(-3) = -15 $

$ a - 9 = -15 \implies a = -15 + 9 = -6 $

Now, finding the sum of the first 10 terms ($ S_{10} $):

\[ S_{10} = \frac{10}{2}[2(-6) + (10 - 1)(-3)] \] \[ S_{10} = 5[-12 + 9(-3)] \] \[ S_{10} = 5[-12 - 27] \] \[ S_{10} = 5(-39) = -195 \]

Final Answer:

The sum of the first 10 numbers is -195.

Question 5

Two given A.P.'s are 9, 7, 5, ... and 24, 21, 18, ... If $ n^{th} $ term of both the progressions are equal then find the value of $ n $ and $ n^{th} $ term.

Solution:

Given: First A.P. is $ 9, 7, 5, \dots $ and Second A.P. is $ 24, 21, 18, \dots $
Condition: The $ n^{th} $ terms of both A.P.'s are equal.
To Find: $ n $ and $ t_n $.

Calculation for First A.P.:

Here, $ a = 9 $, $ d = 7 - 9 = -2 $

$ t_n = a + (n-1)d $

$ t_n = 9 + (n-1)(-2) = 9 - 2n + 2 $

$ t_n = 11 - 2n $ ... (i)

Calculation for Second A.P.:

Here, $ a = 24 $, $ d = 21 - 24 = -3 $

$ t_n = 24 + (n-1)(-3) = 24 - 3n + 3 $

$ t_n = 27 - 3n $ ... (ii)

Equating both $ n^{th} $ terms based on the given condition:

\[ 11 - 2n = 27 - 3n \] \[ 3n - 2n = 27 - 11 \] \[ n = 16 \]

Now, substituting $ n = 16 $ in equation (i) to find the term:

$ t_{16} = 11 - 2(16) = 11 - 32 = -21 $

Final Answer:

The value of $ n $ is 16 and the $ 16^{th} $ term is -21.

Question 6

If sum of $ 3^{rd} $ and $ 8^{th} $ terms of an A.P. is 7 and sum of $ 7^{th} $ and $ 14^{th} $ terms is -3, then find the $ 10^{th} $ term.

Solution:

Given: $ t_3 + t_8 = 7 $ and $ t_7 + t_{14} = -3 $
To Find: The $ 10^{th} $ term ($ t_{10} $)

Calculation:

According to the first condition:

$ t_3 + t_8 = 7 $

$ (a + 2d) + (a + 7d) = 7 $

$ 2a + 9d = 7 $ ... (i)

According to the second condition:

$ t_7 + t_{14} = -3 $

$ (a + 6d) + (a + 13d) = -3 $

$ 2a + 19d = -3 $ ... (ii)

Subtracting equation (i) from (ii):

\[ (2a + 19d) - (2a + 9d) = -3 - 7 \] \[ 10d = -10 \implies d = -1 \]

Substituting $ d = -1 $ in equation (i):

$ 2a + 9(-1) = 7 $

$ 2a - 9 = 7 \implies 2a = 16 \implies a = 8 $

Finding the $ 10^{th} $ term:

\[ t_{10} = a + 9d \] \[ t_{10} = 8 + 9(-1) = 8 - 9 = -1 \]

Final Answer:

The $ 10^{th} $ term of the A.P. is -1.

Question 7

In an A.P. the first term is -5 and last term is 45. If sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?

Solution:

Given: $ a = -5 $, $ t_n = 45 $, and $ S_n = 120 $
To Find: Total terms $ n $ and common difference $ d $.

Memory Trick: When both the first term and the last term are known, use the shortcut formula: $ S_n = \frac{n}{2}(t_1 + t_n) $ to find $ n $ directly!

Formula & Substitution:

\[ S_n = \frac{n}{2}(t_1 + t_n) \] \[ 120 = \frac{n}{2}(-5 + 45) \] \[ 120 = \frac{n}{2}(40) \] \[ 120 = 20n \implies n = 6 \]

Now, finding the common difference $ d $ using $ t_n $:

\[ t_n = a + (n-1)d \] \[ t_6 = -5 + (6 - 1)d \] \[ 45 = -5 + 5d \] \[ 50 = 5d \implies d = 10 \]

Final Answer:

There are 6 terms in the A.P. and the common difference is 10.

Question 8

Sum of 1 to $ n $ natural numbers is 36, then find the value of $ n $.

Solution:

Given: The sequence is $ 1, 2, 3, \dots, n $ (which is an A.P.). $ S_n = 36 $
To Find: The value of $ n $.

Calculation:

Here, $ a = 1 $ and $ d = 1 $.

Formula: $ S_n = \frac{n}{2}[2a + (n-1)d] $

\[ 36 = \frac{n}{2}[2(1) + (n-1)(1)] \] \[ 72 = n[2 + n - 1] \] \[ 72 = n(n + 1) \] \[ n^2 + n - 72 = 0 \]

Factorizing the quadratic equation:

\[ n^2 + 9n - 8n - 72 = 0 \] \[ n(n + 9) - 8(n + 9) = 0 \] \[ (n + 9)(n - 8) = 0 \]

Since $ n $ represents the number of terms, it cannot be negative. Therefore, $ n = -9 $ is discarded.

$ n = 8 $

Final Answer:

The value of $ n $ is 8.

Question 9

Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.

Solution:

Let the three parts be: $ a-d $, $ a $, and $ a+d $ (This is a standard assumption for 3 consecutive terms in A.P.)
Given Conditions: Sum of parts = 207. Product of two smaller parts = 4623.

Calculation:

According to the first condition (Sum):

\[ (a - d) + a + (a + d) = 207 \] \[ 3a = 207 \implies a = 69 \]

According to the second condition (Product of smaller parts):

Assuming $ d > 0 $, the smaller parts are $ (a-d) $ and $ a $.

\[ (a - d) \times a = 4623 \] \[ (69 - d) \times 69 = 4623 \] \[ 69 - d = \frac{4623}{69} \] \[ 69 - d = 67 \] \[ d = 69 - 67 = 2 \]

Now, finding the three parts:

First part = $ a - d = 69 - 2 = 67 $
Second part = $ a = 69 $
Third part = $ a + d = 69 + 2 = 71 $

Final Answer:

The three parts of 207 that are in A.P. are 67, 69, and 71.

Question 10

There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.

Solution:

Given: Total terms $ n = 37 $. Sum of 3 middle terms = 225. Sum of last 3 terms = 429.
To Find: The A.P. sequence.

Calculation:

The middle term of 37 terms is the $ \left(\frac{37+1}{2}\right)^{th} = 19^{th} $ term.

Therefore, the three exact middle terms are $ t_{18}, t_{19}, \text{ and } t_{20} $.

According to the first condition:

\[ t_{18} + t_{19} + t_{20} = 225 \] \[ (a + 17d) + (a + 18d) + (a + 19d) = 225 \] \[ 3a + 54d = 225 \] ... (i)

The last three terms are $ t_{35}, t_{36}, \text{ and } t_{37} $.

According to the second condition:

\[ t_{35} + t_{36} + t_{37} = 429 \] \[ (a + 34d) + (a + 35d) + (a + 36d) = 429 \] \[ 3a + 105d = 429 \] ... (ii)

Subtracting equation (i) from (ii):

\[ (3a + 105d) - (3a + 54d) = 429 - 225 \] \[ 51d = 204 \implies d = 4 \]

Substituting $ d = 4 $ in equation (i):

\[ 3a + 54(4) = 225 \] \[ 3a + 216 = 225 \] \[ 3a = 9 \implies a = 3 \]

Final Answer:

The required A.P. starts with $ a = 3 $ and $ d = 4 $. Therefore, the sequence is 3, 7, 11, 15, ..., 147.

Question 11

If first term of an A.P. is $ a $, second term is $ b $, and last term is $ c $, then show that sum of all terms is $ \frac{(a+c)(b+c-2a)}{2(b-a)} $.

Solution:

Given: $ t_1 = a $, $ t_2 = b $, $ t_n = c $.
To Prove: $ S_n = \frac{(a+c)(b+c-2a)}{2(b-a)} $

Proof:

1. Finding the common difference ($ d $):

\[ d = t_2 - t_1 = b - a \]

2. Expressing total terms ($ n $) in terms of variables:

\[ t_n = a + (n-1)d \] \[ c = a + (n-1)(b-a) \] \[ c - a = (n-1)(b-a) \] \[ n - 1 = \frac{c - a}{b - a} \] \[ n = \frac{c - a}{b - a} + 1 \] \[ n = \frac{c - a + b - a}{b - a} \] \[ n = \frac{b + c - 2a}{b - a} \] ... (i)

3. Finding the Sum ($ S_n $):

\[ S_n = \frac{n}{2}(t_1 + t_n) \]

Substitute the value of $ n $ from equation (i):

\[ S_n = \frac{\frac{b+c-2a}{b-a}}{2}(a+c) \] \[ S_n = \frac{(a+c)(b+c-2a)}{2(b-a)} \]

Hence Proved.

Question 12

If the sum of first $ p $ terms of an A.P. is equal to the sum of first $ q $ terms, then show that the sum of its first $ (p+q) $ terms is zero. ($ p \ne q $)

Solution:

Given: $ S_p = S_q $
To Prove: $ S_{p+q} = 0 $

Proof:

According to the given condition:

\[ S_p = S_q \] \[ \frac{p}{2}[2a + (p-1)d] = \frac{q}{2}[2a + (q-1)d] \]

Multiplying both sides by 2:

\[ p[2a + (p-1)d] = q[2a + (q-1)d] \] \[ 2ap + p(p-1)d = 2aq + q(q-1)d \] \[ 2ap - 2aq + p(p-1)d - q(q-1)d = 0 \] \[ 2a(p - q) + d[p^2 - p - (q^2 - q)] = 0 \] \[ 2a(p - q) + d[p^2 - q^2 - p + q] = 0 \]

Rearranging and factoring:

\[ 2a(p - q) + d[(p-q)(p+q) - (p-q)] = 0 \]

Taking $ (p-q) $ common:

\[ (p-q)[2a + d(p+q-1)] = 0 \]

Since $ p \ne q $, $ p - q \ne 0 $. Therefore:

\[ 2a + (p+q-1)d = 0 \] ... (i)

Now, let's find the sum of $ (p+q) $ terms:

\[ S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d] \]

Substitute from equation (i):

\[ S_{p+q} = \frac{p+q}{2}[0] = 0 \]

Hence Proved.

Question 13

If $ m $ times the $ m^{th} $ term of an A.P. is equal to $ n $ times $ n^{th} $ term, then show that the $ (m+n)^{th} $ term of the A.P. is zero.

Solution:

Given: $ m \cdot t_m = n \cdot t_n $
To Prove: $ t_{m+n} = 0 $

Proof:

\[ m[a + (m-1)d] = n[a + (n-1)d] \] \[ ma + md(m-1) = na + nd(n-1) \] \[ ma - na + m^2d - md - n^2d + nd = 0 \] \[ a(m - n) + d(m^2 - n^2) - d(m - n) = 0 \]

Factoring $ (m^2 - n^2) $:

\[ a(m - n) + d(m+n)(m-n) - d(m - n) = 0 \]

Taking $ (m - n) $ common from the entire equation:

\[ (m - n)[a + (m+n)d - d] = 0 \] \[ (m - n)[a + (m+n-1)d] = 0 \]

Dividing both sides by $ (m - n) $, assuming $ m \ne n $:

\[ a + (m+n-1)d = 0 \]

By definition of the general term:

\[ t_{m+n} = a + (m+n-1)d \] \[ \therefore t_{m+n} = 0 \]

Hence Proved.

Question 14

₹1000 is invested at 10 percent simple interest. Check at the end of every year if the total interest amount is in A.P. If this is an A.P. then find interest amount after 20 years. For this complete the following activity.

Solution & Completed Activity:

Key Formula: $ \text{Simple Interest (I)} = \frac{P \times R \times N}{100} $

Simple interest after $ 1 \text{ year} = \frac{1000 \times 10 \times 1}{100} = 100 $

Simple interest after $ 2 \text{ years} = \frac{1000 \times 10 \times 2}{100} = 200 $

Simple interest after $ 3 \text{ years} = \frac{1000 \times 10 \times 3}{100} = 300 $

According to this, the simple interest for 4, 5, 6 years will be ₹400, ₹500, ₹600 respectively.

From this sequence (100, 200, 300, ...), we can determine:

$ d = 200 - 100 = 100 $

$ a = 100 $

Amount of simple interest after 20 years:

\[ t_n = a + (n-1)d \] \[ t_{20} = 100 + (20-1)100 \] \[ t_{20} = 100 + 19 \times 100 = 100 + 1900 \] \[ t_{20} = 2000 \]

Final Answer:

Amount of simple interest after 20 years is ₹2000.

Maharashtra Board Class 10 Maths Part 1 Chapter 3 Arithmetic Progression Practice Set 3.4 Solutions

Welcome to the complete Textbook Exercise Solutions for Maharashtra Board Class 10 Maths Part 1. In this post, we provide a detailed, step-by-step conceptual guide to solving Practice Set 3.4 from Chapter 3: Arithmetic Progression. These word problems are crucial for board exams, so we have structured every solution with clear Given data, Formulas, and Calculations to help you maximize your scores!

Practice Set 3.4 Questions and Answers

Question 1

On $ 1^{\text{st}} $ Jan 2016, Sanika decides to save ₹10, ₹11 on the second day, ₹12 on the third day. If she decides to save like this, then on $ 31^{\text{st}} $ Dec 2016 what would be her total saving?

Given: Sanika's daily savings form an Arithmetic Progression (A.P.): $ 10, 11, 12, \dots $
First term $ a = 10 $
Common difference $ d = 11 - 10 = 1 $
Total number of days $ n = 366 $ (since 2016 is a leap year).
Formula:
Key Formula (Sum of an A.P.):
$ S_{n} = \frac{n}{2}[2a + (n-1)d] $
Substitution & Calculation: \[ \begin{aligned} S_{366} &= \frac{366}{2}[2(10) + (366 - 1)1] \\ &= 183(20 + 365 \times 1) \\ &= 183(20 + 365) \\ &= 183 \times 385 \\ S_{366} &= 70455 \end{aligned} \]
Final Answer: Sanika's total saving on $ 31^{\text{st}} $ December 2016 would be ₹70,455.

Question 2

A man borrows ₹8000 and agrees to repay with a total interest of ₹1360 in 12 monthly instalments. Each instalment being less than the preceding one by ₹40. Find the amount of the first and last instalment.

Given: The instalments are in an A.P.
Total amount repaid in 12 instalments ($ S_{12} $) = Amount borrowed + Total interest = $ 8000 + 1360 = 9360 $.
Number of instalments $ n = 12 $.
Each instalment is less than the preceding one by ₹40, therefore $ d = -40 $.
Formula:
Formulas Used:
Sum formula: $ S_{n} = \frac{n}{2}[2a + (n-1)d] $
$ n^{\text{th}} $ term formula: $ t_{n} = a + (n-1)d $
Substitution & Calculation: Finding the first instalment ($ a $): \[ \begin{aligned} S_{12} &= \frac{12}{2}[2a + (12 - 1)(-40)] \\ 9360 &= 6[2a + (11)(-40)] \\ 9360 &= 6(2a - 440) \\ \frac{9360}{6} &= 2a - 440 \\ 1560 &= 2a - 440 \\ 1560 + 440 &= 2a \\ 2000 &= 2a \\ a &= \frac{2000}{2} \\ a &= 1000 \end{aligned} \] Finding the last instalment ($ t_{12} $): \[ \begin{aligned} t_{12} &= 1000 + (12 - 1)(-40) \\ &= 1000 + 11(-40) \\ &= 1000 - 440 \\ t_{12} &= 560 \end{aligned} \]
Final Answer: The amount of the first instalment is ₹1000 and that of the last instalment is ₹560.

Question 3

Sachin invested in a national saving certificate scheme. In the first year he invested ₹5000, in the second year ₹7000, in the third year ₹9000 and so on. Find the total amount that he invested in 12 years.

Given: Amounts invested by Sachin each year: $ 5000, 7000, 9000, \dots $
This sequence is an A.P.
First term $ a = 5000 $
Common difference $ d = 7000 - 5000 = 2000 $
Number of years $ n = 12 $.
Formula: $ S_{n} = \frac{n}{2}[2a + (n-1)d] $
Substitution & Calculation: \[ \begin{aligned} S_{12} &= \frac{12}{2}[2(5000) + (12 - 1)2000] \\ &= 6(10000 + 11 \times 2000) \\ &= 6(10000 + 22000) \\ &= 6(32000) \\ S_{12} &= 192000 \end{aligned} \]
Final Answer: The total amount invested by Sachin in 12 years is ₹1,92,000.

Question 4

There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the $ 15^{\text{th}} $ row and also find how many total seats are there in the auditorium?

Given: The number of seats arranged row-wise are: $ 20, 22, 24, \dots $
This is an A.P. where:
First term $ a = 20 $
Common difference $ d = 22 - 20 = 2 $
Total rows $ n = 27 $.
Formula:
$ n^{\text{th}} $ term: $ t_{n} = a + (n-1)d $
Sum of n terms: $ S_{n} = \frac{n}{2}[2a + (n-1)d] $
Substitution & Calculation: Finding seats in the $ 15^{\text{th}} $ row ($ t_{15} $): \[ \begin{aligned} t_{15} &= 20 + (15 - 1)2 \\ &= 20 + 14 \times 2 \\ &= 20 + 28 \\ t_{15} &= 48 \end{aligned} \] Finding total seats ($ S_{27} $): \[ \begin{aligned} S_{27} &= \frac{27}{2}[2(20) + (27 - 1)2] \\ &= \frac{27}{2}(40 + 26 \times 2) \\ &= \frac{27}{2}(40 + 52) \\ &= \frac{27}{2} \times 92 \\ &= 27 \times 46 \\ S_{27} &= 1242 \end{aligned} \]
Final Answer: The number of seats in the $ 15^{\text{th}} $ row is 48 and the total seats in the auditorium are 1242.

Question 5

Kargil's temperature was recorded in a week from Monday to Saturday. All readings were in A.P. The sum of temperatures of Monday and Saturday was $ 5^{\circ}\text{C} $ more than sum of temperatures of Tuesday and Saturday. If temperature of Wednesday was $ -30^{\circ}\text{C} $ then find the temperature on the other five days.

Given: Let the temperatures from Monday to Saturday in A.P. be:
$ a,\ a+d,\ a+2d,\ a+3d,\ a+4d,\ a+5d $
Condition 1: Sum of Monday & Saturday = Sum of Tuesday & Saturday + $ 5^{\circ} $
Condition 2: Temperature on Wednesday = $ -30^{\circ}\text{C} $
Substitution & Calculation: From the first condition: \[ \begin{aligned} (a) + (a+5d) &= (a+d) + (a+5d) + 5 \\ 2a + 5d &= 2a + 6d + 5 \\ 5d - 6d &= 5 \\ -d &= 5 \\ d &= -5^{\circ} \end{aligned} \] From the second condition (Wednesday is the $ 3^{\text{rd}} $ term): \[ \begin{aligned} a + 2d &= -30^{\circ} \\ a + 2(-5^{\circ}) &= -30^{\circ} \\ a - 10^{\circ} &= -30^{\circ} \\ a &= -30^{\circ} + 10^{\circ} \\ a &= -20^{\circ} \end{aligned} \] Finding the temperatures for all days:
- Monday ($ a $): $ -20^{\circ}\text{C} $
- Tuesday ($ a+d $): $ -20^{\circ} - 5^{\circ} = -25^{\circ}\text{C} $
- Wednesday ($ a+2d $): $ -30^{\circ}\text{C} $ (Given)
- Thursday ($ a+3d $): $ -20^{\circ} + 3(-5^{\circ}) = -20^{\circ} - 15^{\circ} = -35^{\circ}\text{C} $
- Friday ($ a+4d $): $ -20^{\circ} + 4(-5^{\circ}) = -20^{\circ} - 20^{\circ} = -40^{\circ}\text{C} $
- Saturday ($ a+5d $): $ -20^{\circ} + 5(-5^{\circ}) = -20^{\circ} - 25^{\circ} = -45^{\circ}\text{C} $
Final Answer: The temperatures on the other five days are $ -20^{\circ}\text{C} $, $ -25^{\circ}\text{C} $, $ -35^{\circ}\text{C} $, $ -40^{\circ}\text{C} $, and $ -45^{\circ}\text{C} $.

Question 6

On the world environment day, a tree plantation programme was arranged on a land which is triangular in shape. Trees are planted such that in the first row there is one tree, in the second row there are two trees, in the third row three trees and so on. Find the total number of trees in the 25 rows.

Given: The number of trees planted row-wise are: $ 1, 2, 3, \dots $
This sequence is an A.P.
First term $ a = 1 $
Common difference $ d = 2 - 1 = 1 $
Total number of rows $ n = 25 $.
Formula: $ S_{n} = \frac{n}{2}[2a + (n-1)d] $
Substitution & Calculation: \[ \begin{aligned} S_{25} &= \frac{25}{2}[2(1) + (25 - 1)1] \\ &= \frac{25}{2}(2 + 24) \\ &= \frac{25}{2} \times 26 \\ &= 25 \times 13 \\ S_{25} &= 325 \end{aligned} \]
Final Answer: The total number of trees planted in the 25 rows is 325.

Maharashtra Board Class 10 Maths Part 1 Practice Set 3.3 Solutions

Welcome to the comprehensive Textbook Exercise Solutions for Maharashtra Board Class 10 Maths Part 1 (Algebra). In this post, we provide complete, step-by-step answers for Chapter 3: Arithmetic Progression, Practice Set 3.3. Designed to help you prepare effectively for your SSC board exams, these solutions break down complex numericals, highlight key formulas, and provide clear calculation steps to ensure you score full marks!

Practice Set 3.3 Solutions

Question 1

First term and common difference of an A.P. are 6 and 3 respectively; find $ S_{27} $.

Given:

First term $ (a) = 6 $
Common difference $ (d) = 3 $
$ S_{27} = ? $

Key Formula: \[ S_{n} = \frac{n}{2}[2a + (n-1)d] \]

Calculation:

Substitute the given values into the formula:
$ S_{27} = \frac{27}{2}[2(6) + (27 - 1)3] $
$ S_{27} = \frac{27}{2}[12 + (26 \times 3)] $
$ S_{27} = \frac{27}{2}[12 + 78] $
$ S_{27} = \frac{27}{2} \times 90 $
$ S_{27} = 27 \times 45 $
$ S_{27} = 1215 $

Final Answer: The value of $ S_{27} $ is 1215.

Question 2

Find the sum of the first 123 even natural numbers.

Solution:

The sequence of even natural numbers is: 2, 4, 6, 8, ...
This sequence forms an Arithmetic Progression (A.P.).

Given:

First term $ (a) = 2 $
Common difference $ (d) = 4 - 2 = 2 $
Number of terms $ (n) = 123 $

Calculation:

$ S_{n} = \frac{n}{2}[2a + (n-1)d] $
$ S_{123} = \frac{123}{2}[2(2) + (123 - 1)2] $
$ S_{123} = \frac{123}{2}[4 + 122(2)] $
$ S_{123} = \frac{123}{2}[4 + 244] $
$ S_{123} = \frac{123}{2} \times 248 $
$ S_{123} = 123 \times 124 $
$ S_{123} = 15252 $

Final Answer: The sum of the first 123 even natural numbers is 15252.

Question 3

Find the sum of all even numbers between 1 and 350.

Solution:

The even numbers strictly between 1 and 350 are: 2, 4, 6, ..., 348.
This sequence forms an A.P.

Given:

First term $ (a) = 2 $
Common difference $ (d) = 4 - 2 = 2 $
Last term $ (t_{n}) = 348 $

Exam Tip: First, find the total number of terms $ (n) $ using the $ t_{n} $ formula before calculating the sum.

Step 1: Find $ n $

$ t_{n} = a + (n-1)d $
$ 348 = 2 + (n-1)2 $
$ 348 - 2 = (n-1)2 $
$ 346 = (n-1)2 $
$ n - 1 = \frac{346}{2} $
$ n - 1 = 173 $
$ n = 173 + 1 = 174 $

Step 2: Calculate Sum $ (S_{n}) $

$ S_{n} = \frac{n}{2}[2a + (n-1)d] $
$ S_{174} = \frac{174}{2}[2(2) + (174 - 1)2] $
$ S_{174} = 87[4 + 173 \times 2] $
$ S_{174} = 87[4 + 346] $
$ S_{174} = 87 \times 350 $
$ S_{174} = 30450 $

Final Answer: The sum of all even numbers between 1 and 350 is 30450.

Question 4

In an A.P., the 19^th term is 52 and the 38^th term is 128. Find the sum of the first 56 terms.

Given:

$ t_{19} = 52 $
$ t_{38} = 128 $

Calculation:

Let $ a $ be the first term and $ d $ be the common difference.

Using the formula $ t_{n} = a + (n-1)d $:
$ t_{19} = a + (19 - 1)d $
$ 52 = a + 18d \quad \text{... (i)} $

Similarly, for the 38^th term:
$ t_{38} = a + (38 - 1)d $
$ 128 = a + 37d \quad \text{... (ii)} $

Adding equations (i) and (ii):

$ (a + 18d) + (a + 37d) = 52 + 128 $
$ 2a + 55d = 180 \quad \text{... (iii)} $

Finding $ S_{56} $:

$ S_{n} = \frac{n}{2}[2a + (n-1)d] $
$ S_{56} = \frac{56}{2}[2a + (56 - 1)d] $
$ S_{56} = 28(2a + 55d) $
Substituting the value from equation (iii):
$ S_{56} = 28 \times 180 $
$ S_{56} = 5040 $

Final Answer: The sum of the first 56 terms is 5040.

Question 5

Complete the following activity to find the sum of natural numbers between 1 to 140 which are divisible by 4.

Solution (Activity Completed):

From 1 to 140, natural numbers divisible by 4 are:
$ 4, 8, 12, \dots, 136 $

How many numbers? $ n = \mathbf{34} $
$ n = \mathbf{34}, \ a = \mathbf{4}, \ d = \mathbf{4} $

Step 1: Finding $ n $
$ t_{n} = a + (n-1)d $
$ 136 = \mathbf{4} + (n-1) \times \mathbf{4} $
$ n = \mathbf{34} $

Step 2: Finding Sum $ S_{n} $
$ S_{n} = \frac{n}{2}[2a + (n-1)d] $
$ S_{\mathbf{34}} = \frac{\mathbf{34}}{2}[2 \times 4 + (34 - 1)4] $
$ S_{34} = 17[8 + 132] = \mathbf{2380} $

Final Answer: The sum of numbers from 1 to 140 which are divisible by 4 is 2380.

Question 6

Sum of the first 55 terms in an A.P. is 3300. Find its 28^th term.

Given:

$ S_{55} = 3300 $

Calculation:

$ S_{n} = \frac{n}{2}[2a + (n-1)d] $
$ S_{55} = \frac{55}{2}[2a + (55 - 1)d] $
$ 3300 = \frac{55}{2}[2a + 54d] $
Taking 2 common from the bracket:
$ 3300 = \frac{55}{2} \times 2(a + 27d) $
$ 3300 = 55(a + 27d) $
$ a + 27d = \frac{3300}{55} $
$ a + 27d = 60 \quad \text{... (i)} $

Finding the 28^th term:

$ t_{n} = a + (n-1)d $
$ t_{28} = a + (28 - 1)d $
$ t_{28} = a + 27d $
From equation (i):
$ t_{28} = 60 $

Final Answer: The 28^th term of the A.P. is 60.

Question 7

In an A.P., the sum of three consecutive terms is 27 and their product is 504. Find the terms.
(Assume that three consecutive terms in A.P. are $ a - d $, $ a $, and $ a + d $.)

Solution:

Let the three consecutive terms in an A.P. be $ a - d $, $ a $, and $ a + d $.

First Condition (Sum is 27):

$ (a - d) + a + (a + d) = 27 $
$ 3a = 27 $
$ a = \frac{27}{3} $
$ a = 9 \quad \text{... (i)} $

Second Condition (Product is 504):

$ (a - d) \times a \times (a + d) = 504 $
$ a(a^{2} - d^{2}) = 504 $
Substitute $ a = 9 $ from (i):
$ 9(9^{2} - d^{2}) = 504 $
$ 9(81 - d^{2}) = 504 $
$ 81 - d^{2} = \frac{504}{9} $
$ 81 - d^{2} = 56 $
$ d^{2} = 81 - 56 $
$ d^{2} = 25 $
Taking square root on both sides:
$ d = \pm 5 $

Case 1: When $ a = 9 $ and $ d = 5 $

$ a - d = 9 - 5 = 4 $
$ a = 9 $
$ a + d = 9 + 5 = 14 $

Case 2: When $ a = 9 $ and $ d = -5 $

$ a - d = 9 - (-5) = 9 + 5 = 14 $
$ a = 9 $
$ a + d = 9 - 5 = 4 $

Final Answer: The three consecutive terms are 4, 9, 14 or 14, 9, 4.

Question 8

Find four consecutive terms in an A.P. whose sum is 12 and the sum of the 3^rd and 4^th term is 14.
(Assume the four consecutive terms in A.P. are $ a - d $, $ a $, $ a + d $, and $ a + 2d $.)

Solution:

Let the four consecutive terms be $ a - d $, $ a $, $ a + d $, and $ a + 2d $.

First Condition (Sum is 12):

$ (a - d) + a + (a + d) + (a + 2d) = 12 $
$ 4a + 2d = 12 $
Dividing by 2:
$ 2a + d = 6 \quad \text{... (i)} $

Second Condition (Sum of 3^rd and 4^th terms is 14):

$ (a + d) + (a + 2d) = 14 $
$ 2a + 3d = 14 \quad \text{... (ii)} $

Subtracting equation (i) from (ii):

$ (2a + 3d) - (2a + d) = 14 - 6 $
$ 2d = 8 $
$ d = \frac{8}{2} = 4 $

Substitute $ d = 4 $ in equation (i):

$ 2a + 4 = 6 $
$ 2a = 6 - 4 = 2 $
$ a = \frac{2}{2} = 1 $

Finding the terms:

$ a - d = 1 - 4 = -3 $
$ a = 1 $
$ a + d = 1 + 4 = 5 $
$ a + 2d = 1 + 2(4) = 1 + 8 = 9 $

Final Answer: The four consecutive terms are -3, 1, 5, and 9.

Question 9

If the 9^th term of an A.P. is zero, then show that the 29^th term is twice the 19^th term.

To Prove: $ t_{29} = 2 \times t_{19} $

Proof:

Let $ a $ be the first term and $ d $ be the common difference.

Given: $ t_{9} = 0 $
$ t_{n} = a + (n-1)d $
$ t_{9} = a + (9 - 1)d $
$ 0 = a + 8d $
$ a = -8d \quad \text{... (i)} $

Finding the 19^th term:
$ t_{19} = a + (19 - 1)d $
$ t_{19} = a + 18d $
Substitute $ a = -8d $ from (i):
$ t_{19} = -8d + 18d $
$ t_{19} = 10d \quad \text{... (ii)} $

Finding the 29^th term:
$ t_{29} = a + (29 - 1)d $
$ t_{29} = a + 28d $
Substitute $ a = -8d $ from (i):
$ t_{29} = -8d + 28d $
$ t_{29} = 20d $

Conclusion:
We can write $ 20d $ as $ 2(10d) $.
From equation (ii), we know $ 10d = t_{19} $.
Therefore, $ t_{29} = 2(t_{19}) $.

Hence Proved: The 29^th term is twice the 19^th term.

Additional Question (Textbook Pg. No. 71)

Find the sum of all odd numbers from 1 to 150.

Solution:

Odd numbers from 1 to 150 are: 1, 3, 5, 7, ..., 149.
Here, the difference between any two consecutive terms is 2, hence it forms an A.P.

Given:

First term $ (a) = 1 $
Common difference $ (d) = 2 $
Last term $ (t_{n}) = 149 $

Step 1: Find the number of terms $ (n) $

$ t_{n} = a + (n-1)d $
$ 149 = 1 + (n-1)2 $
$ 149 - 1 = (n-1)2 $
$ \frac{148}{2} = n - 1 $
$ 74 = n - 1 $
$ n = 75 $

Step 2: Find the sum $ (S_{75}) $

Memory Trick: If both the first and last terms are known, you can use the shortcut formula: $ S_{n} = \frac{n}{2}(t_{1} + t_{n}) $.

Method I (Using standard formula):
$ S_{n} = \frac{n}{2}[2a + (n-1)d] $
$ S_{75} = \frac{75}{2}[2(1) + (75 - 1)2] $
$ S_{75} = \frac{75}{2}[2 + 148] $
$ S_{75} = \frac{75}{2}[150] $
$ S_{75} = 75 \times 75 = 5625 $

Method II (Using shortcut formula):
$ S_{n} = \frac{n}{2}(t_{1} + t_{n}) $
$ S_{75} = \frac{75}{2}(1 + 149) $
$ S_{75} = \frac{75}{2}(150) $
$ S_{75} = 75 \times 75 = 5625 $

Final Answer: The sum of all odd numbers from 1 to 150 is 5625.

Class 10 Maths Part 1 Chapter 3 Arithmetic Progression Practice Set 3.2 Solutions (Maharashtra Board)

Welcome to our comprehensive guide for Maharashtra Board Class 10 Maths Part 1. In this post, we provide complete, step-by-step Textbook Exercise Solutions for Chapter 3: Arithmetic Progression, specifically focusing on Practice Set 3.2. These verified solutions will help you master sequence calculations, score maximum marks in your board exams, and easily find the nth term of any sequence.

Question 1: Write the correct number in the given boxes from the following A.P.

Given A.P.: $ 1, 8, 15, 22, \dots $[cite: 1]

Calculation:
- Here $ a = 1 $, $ t_1 = 1 $, $ t_2 = 8 $, $ t_3 = 15 $[cite: 2]
- $ t_2 - t_1 = 8 - 1 = 7 $[cite: 2]
- $ t_3 - t_2 = 15 - 8 = 7 $[cite: 2]
Final Answer: $ d = 7 $[cite: 2]
Given A.P.: $ 3, 6, 9, 12, \dots $[cite: 1]

Calculation:
- Here $ t_1 = 3 $, $ t_2 = 6 $, $ t_3 = 9 $, $ t_4 = 12 $[cite: 2]
- $ t_2 - t_1 = 6 - 3 = 3 $ (Note: Derived from sequence pattern)
- $ t_3 - t_2 = 9 - 6 = 3 $[cite: 2]
Final Answer: $ d = 3 $[cite: 2]
Given A.P.: $ -3, -8, -13, -18, \dots $[cite: 1]

Calculation:
- Here $ t_1 = -3 $, $ t_2 = -8 $, $ t_3 = -13 $, $ t_4 = -18 $[cite: 2]
- $ t_2 - t_1 = -8 - (-3) = -5 $[cite: 2]
- $ t_3 - t_2 = -13 - (-8) = -5 $[cite: 2]
Final Answer: $ d = -5 $[cite: 2]
Given A.P.: $ 70, 60, 50, 40, \dots $[cite: 1]

Calculation:
- Here $ t_1 = 70 $, $ t_2 = 60 $, $ t_3 = 50 $[cite: 2]
- $ a = 70 $[cite: 2]
- $ t_2 - t_1 = 60 - 70 = -10 $
Final Answer: $ d = -10 $[cite: 2]

Question 2: Decide whether following sequence is an A.P., if so find the 20th term of the progression.

Given Sequence: $ -12, -5, 2, 9, 16, 23, 30, \dots $[cite: 2]

Calculation (Checking for A.P.):

Here, $ t_1 = -12 $, $ t_2 = -5 $, $ t_3 = 2 $, $ t_4 = 9 $[cite: 2]
$ t_2 - t_1 = -5 - (-12) = -5 + 12 = 7 $[cite: 2]
$ t_3 - t_2 = 2 - (-5) = 2 + 5 = 7 $[cite: 2]
$ t_4 - t_3 = 9 - 2 = 7 $[cite: 2]
$ t_2 - t_1 = t_3 - t_2 = \dots = 7 = d = \text{constant} $[cite: 2]

The difference between two consecutive terms is constant. Therefore, the given sequence is an Arithmetic Progression (A.P.).[cite: 3]

Key Formula: $ t_n = a + (n-1)d $[cite: 3]

Calculation (Finding 20th term):

Substitute $ a = -12 $, $ d = 7 $, and $ n = 20 $[cite: 3]
$ t_{20} = -12 + (20 - 1)7 $[cite: 3]
$ t_{20} = -12 + 19 \times 7 $[cite: 3]
$ t_{20} = -12 + 133 $[cite: 3]
$ t_{20} = 121 $[cite: 3]

Final Answer: The 20th term of the given A.P. is 121.[cite: 3]

Question 3: Given Arithmetic Progression is 12, 16, 20, 24, ... Find the 24th term of this progression.

Given: A.P. is $ 12, 16, 20, 24, \dots $[cite: 3]

First term $ a = 12 $[cite: 3]
Common difference $ d = 16 - 12 = 4 $[cite: 3]

Formula: $ t_n = a + (n-1)d $[cite: 3]

Calculation:

$ t_{24} = 12 + (24 - 1)4 $[cite: 3]
$ t_{24} = 12 + 23 \times 4 $[cite: 3]
$ t_{24} = 12 + 92 $[cite: 3]
$ t_{24} = 104 $[cite: 3]

Final Answer: The 24th term of the given A.P. is 104.[cite: 3]

Question 4: Find the 19th term of the following A.P. 7, 13, 19, 25, ...

Given: A.P. is $ 7, 13, 19, 25, \dots $[cite: 3]

First term $ a = 7 $[cite: 3]
Common difference $ d = 13 - 7 = 6 $[cite: 3]

Formula: $ t_n = a + (n-1)d $[cite: 3]

Calculation:

$ t_{19} = 7 + (19 - 1)6 $[cite: 3]
$ t_{19} = 7 + 18 \times 6 $[cite: 3]
$ t_{19} = 7 + 108 $[cite: 3]
$ t_{19} = 115 $[cite: 4]

Final Answer: The 19th term of the given A.P. is 115.[cite: 4]

Question 5: Find the 27th term of the following A.P. 9, 4, -1, -6, -11, ...

Given: A.P. is $ 9, 4, -1, -6, -11, \dots $[cite: 4]

First term $ a = 9 $[cite: 4]
Common difference $ d = 4 - 9 = -5 $[cite: 4]

Formula: $ t_n = a + (n-1)d $[cite: 4]

Calculation:

$ t_{27} = 9 + (27 - 1)(-5) $[cite: 4]
$ t_{27} = 9 + 26 \times (-5) $[cite: 4]
$ t_{27} = 9 - 130 $[cite: 4]
$ t_{27} = -121 $[cite: 4]

Final Answer: The 27th term of the given A.P. is -121.[cite: 4]

Question 6: Find how many three-digit natural numbers are divisible by 5.

Given:

The three-digit natural numbers divisible by 5 are $ 100, 105, 110, \dots, 995 $.[cite: 4]
This sequence is an A.P.[cite: 4]
First term $ a = 100 $[cite: 4]
Common difference $ d = 105 - 100 = 5 $[cite: 4]
Last term $ t_n = 995 $[cite: 4]

Formula: $ t_n = a + (n-1)d $[cite: 4]

Calculation:

$ 995 = 100 + (n - 1)5 $[cite: 4]
$ 995 - 100 = (n - 1)5 $[cite: 4]
$ 895 = (n - 1)5 $[cite: 4]
$ n - 1 = \frac{895}{5} $[cite: 4]
$ n - 1 = 179 $[cite: 5]
$ n = 179 + 1 = 180 $[cite: 5]

Final Answer: There are 180 three-digit natural numbers which are divisible by 5.[cite: 5]

Exam Tip: Whenever asked for "how many numbers", form the A.P., identify the first and last term, and solve for $ n $ using the $ t_n $ formula.

Question 7: The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41st term of that A.P.

Given:

$ t_{11} = 16 $[cite: 5]
$ t_{21} = 29 $[cite: 5]

Formula: $ t_n = a + (n-1)d $[cite: 5]

Calculation:

For the 11th term: $ t_{11} = a + (11 - 1)d $ → $ 16 = a + 10d $. Let this be Equation (i): $ a + 10d = 16 $.[cite: 5]
For the 21st term: $ t_{21} = a + (21 - 1)d $ → $ 29 = a + 20d $. Let this be Equation (ii): $ a + 20d = 29 $.[cite: 5]
Subtracting equation (i) from (ii):
$ (a + 20d) - (a + 10d) = 29 - 16 $
$ 10d = 13 $[cite: 6]
$ d = \frac{13}{10} $[cite: 6]
Substituting $ d = \frac{13}{10} $ in equation (i):
$ a + 10(\frac{13}{10}) = 16 $[cite: 6]
$ a + 13 = 16 $[cite: 6]
$ a = 16 - 13 = 3 $[cite: 6]
Now, to find the 41st term:
$ t_{41} = 3 + (41 - 1)(\frac{13}{10}) $[cite: 6]
$ t_{41} = 3 + 40 \times \frac{13}{10} $[cite: 6]
$ t_{41} = 3 + 52 $[cite: 6]
$ t_{41} = 55 $[cite: 6]

Final Answer: The 41st term of the A.P. is 55.[cite: 6]

Question 8: 11, 8, 5, 2, ... In this A.P. which term is number -151?

Given:

The A.P. is $ 11, 8, 5, 2, \dots $[cite: 6]
$ a = 11 $[cite: 6]
$ d = 8 - 11 = -3 $[cite: 6]
Let the nth term be -151, so $ t_n = -151 $[cite: 6]

Formula: $ t_n = a + (n-1)d $[cite: 6]

Calculation:

$ -151 = 11 + (n - 1)(-3) $[cite: 6]
$ -151 - 11 = (n - 1)(-3) $[cite: 6]
$ -162 = (n - 1)(-3) $[cite: 6]
$ n - 1 = \frac{-162}{-3} $[cite: 6, 7]
$ n - 1 = 54 $[cite: 7]
$ n = 54 + 1 = 55 $[cite: 7]

Final Answer: The 55th term of the given A.P. is -151.[cite: 7]

Question 9: In the natural numbers from 10 to 250, how many are divisible by 4?

Given:

The natural numbers from 10 to 250 divisible by 4 form the sequence: $ 12, 16, 20, \dots, 248 $.[cite: 7]
This sequence is an A.P.[cite: 7]
$ a = 12 $[cite: 7]
$ d = 16 - 12 = 4 $[cite: 7]
Let the number of terms be $ n $, so $ t_n = 248 $.[cite: 7]

Formula: $ t_n = a + (n-1)d $[cite: 7]

Calculation:

$ 248 = 12 + (n - 1)4 $[cite: 7]
$ 248 - 12 = (n - 1)4 $[cite: 7]
$ 236 = (n - 1)4 $[cite: 7]
$ n - 1 = \frac{236}{4} $[cite: 7]
$ n - 1 = 59 $[cite: 7]
$ n = 59 + 1 = 60 $[cite: 7]

Final Answer: There are 60 natural numbers from 10 to 250 which are divisible by 4.[cite: 7]

Question 10: In an A.P., the 17th term is 7 more than its 10th term. Find the common difference.

Given Condition: $ t_{17} = t_{10} + 7 $[cite: 8]

Formula: $ t_n = a + (n-1)d $[cite: 8]

Calculation:

$ a + (17 - 1)d = a + (10 - 1)d + 7 $[cite: 8]
$ a + 16d = a + 9d + 7 $[cite: 8]
$ a + 16d - a - 9d = 7 $[cite: 8]
$ 7d = 7 $[cite: 8]
$ d = \frac{7}{7} = 1 $[cite: 8]

Final Answer: The common difference is 1.[cite: 8]

Application Question 1: Kabir's Height

Problem Statement: Kabir's mother keeps a record of his height on each birthday. When he was one year old, his height was 70 cm, at 2 years he was 80 cm tall and 3 years he was 90 cm tall. His aunt Meera was studying in the 10th class. She said, "it seems like Kabir's height grows in Arithmetic Progression". Assuming this, she calculated how tall Kabir will be at the age of 15 years when he is in 10th. You too assume that Kabir grows in A.P. and find out his height at the age of 15 years.[cite: 8]

Given:

Height at 1 year = 70 cm ($ a = 70 $)[cite: 8]
Height at 2 years = 80 cm[cite: 8]
Height at 3 years = 90 cm[cite: 8]
The heights form an A.P. Common difference $ d = 80 - 70 = 10 $.[cite: 8]
We need to find his height at age 15 ($ t_{15} $).[cite: 8]

Formula: $ t_n = a + (n-1)d $[cite: 8]

Calculation:

$ t_{15} = 70 + (15 - 1)10 $[cite: 8]
$ t_{15} = 70 + 14 \times 10 = 70 + 140 $[cite: 8]
$ t_{15} = 210 $[cite: 9]

Final Answer: The height of Kabir at the age of 15 years will be 210 cm.[cite: 9]

Application Question 2: Testing Sequence Terms

Problem Statement: Is 5, 8, 11, 14, ... an A.P.? If so then what will be the 100th term? Check whether 92 is in this A.P.? Is number 61 in this A.P.?[cite: 9]

Part i: Is it an A.P.?

Given: $ 5, 8, 11, 14, \dots $[cite: 9]
$ t_1 = 5 $, $ t_2 = 8 $, $ t_3 = 11 $, $ t_4 = 14 $[cite: 9]
$ t_2 - t_1 = 8 - 5 = 3 $[cite: 9]
$ t_3 - t_2 = 11 - 8 = 3 $[cite: 9]
$ t_4 - t_3 = 14 - 11 = 3 $[cite: 9]

The difference between two consecutive terms is constant ($ d = 3 $). Therefore, the given sequence is an A.P.[cite: 9]

Part ii: What will be the 100th term?

$ t_{100} = 5 + (100 - 1)3 $ (using $ a = 5 $, $ d = 3 $)[cite: 9]
$ t_{100} = 5 + 99 \times 3 $[cite: 9]
$ t_{100} = 5 + 297 $[cite: 9]
$ t_{100} = 302 $[cite: 9]

The 100th term is 302.[cite: 9]

Part iii: Check whether 92 is in this A.P.

Let $ t_n = 92 $[cite: 9]
$ 92 = 5 + (n - 1)3 $[cite: 9]
$ 92 = 5 + 3n - 3 $[cite: 9]
$ 92 = 2 + 3n $[cite: 9]
$ 90 = 3n $[cite: 9]
$ n = \frac{90}{3} = 30 $[cite: 9, 10]

Since $ n $ is a natural number, 92 is the 30th term of the given A.P.[cite: 10]

Part iv: Check whether 61 is in this A.P.

Let $ t_n = 61 $[cite: 10]
$ 61 = 5 + (n - 1)3 $[cite: 10]
$ 61 = 5 + 3n - 3 $[cite: 10]
$ 61 = 2 + 3n $[cite: 10]
$ 61 - 2 = 3n $[cite: 10]
$ 59 = 3n $[cite: 10]
$ n = \frac{59}{3} $[cite: 10]

But, $ n $ must be a natural number, and 59 is not divisible by 3. Therefore, 61 is not in the given A.P.[cite: 10]

Class 10 Maths Part 1 Practice Set 3.1 Solutions | Maharashtra Board

Welcome to our comprehensive guide for Class 10 Maths Part 1 Practice Set 3.1. If you are a Maharashtra Board student studying Chapter 3 Arithmetic Progression, these step-by-step Textbook Exercise Solutions will help you master the core concepts. From identifying whether a sequence is an Arithmetic Progression (A.P.) to calculating the first term and common difference, our detailed breakdown is designed to help you secure full marks in your board exams. Let's dive into the solutions!

Exam Tip: To prove a sequence is an A.P., always show that the difference between at least three consecutive pairs of terms, i.e., $ t_2 - t_1 $ and $ t_3 - t_2 $, is constant. Always enclose negative numbers in brackets during subtraction to avoid careless sign errors!

Practice Set 3.1: Question 1

Which of the following sequences are A.P.? If they are A.P. find the common difference.

i. $ 2, 4, 6, 8, \dots $

Given Sequence: $ 2, 4, 6, 8, \dots $
Here: $ t_1 = 2 $, $ t_2 = 4 $, $ t_3 = 6 $, $ t_4 = 8 $
Calculation:
$ t_2 - t_1 = 4 - 2 = 2 $
$ t_3 - t_2 = 6 - 4 = 2 $
$ t_4 - t_3 = 8 - 6 = 2 $
Since $ t_2 - t_1 = t_3 - t_2 = \dots = 2 = d $ (constant).
The difference between two consecutive terms is constant.
Final Answer: The given sequence is an A.P. and the common difference $ (d) = 2 $.

ii. $ 2, \frac{5}{2}, 3, \frac{7}{2}, \dots $

Given Sequence: $ 2, \frac{5}{2}, 3, \frac{7}{2}, \dots $
Here: $ t_1 = 2 $, $ t_2 = \frac{5}{2} $, $ t_3 = 3 $, $ t_4 = \frac{7}{2} $
Calculation:
$ t_2 - t_1 = \frac{5}{2} - 2 = \frac{5 - 4}{2} = \frac{1}{2} $
$ t_3 - t_2 = 3 - \frac{5}{2} = \frac{6 - 5}{2} = \frac{1}{2} $
$ t_4 - t_3 = \frac{7}{2} - 3 = \frac{7 - 6}{2} = \frac{1}{2} $
Since $ t_2 - t_1 = t_3 - t_2 = \dots = \frac{1}{2} = d $ (constant).
The difference between two consecutive terms is constant.
Final Answer: The given sequence is an A.P. and the common difference $ (d) = \frac{1}{2} $.

iii. $ -10, -6, -2, 2, \dots $

Given Sequence: $ -10, -6, -2, 2, \dots $
Here: $ t_1 = -10 $, $ t_2 = -6 $, $ t_3 = -2 $, $ t_4 = 2 $
Calculation:
$ t_2 - t_1 = -6 - (-10) = -6 + 10 = 4 $
$ t_3 - t_2 = -2 - (-6) = -2 + 6 = 4 $
$ t_4 - t_3 = 2 - (-2) = 2 + 2 = 4 $
Since $ t_2 - t_1 = t_3 - t_2 = \dots = 4 = d $ (constant).
The difference between two consecutive terms is constant.
Final Answer: The given sequence is an A.P. and the common difference $ (d) = 4 $.

iv. $ 0.3, 0.33, 0.333, \dots $

Given Sequence: $ 0.3, 0.33, 0.333, \dots $
Here: $ t_1 = 0.3 $, $ t_2 = 0.33 $, $ t_3 = 0.333 $
Calculation:
$ t_2 - t_1 = 0.33 - 0.3 = 0.03 $
$ t_3 - t_2 = 0.333 - 0.33 = 0.003 $
Since $ t_2 - t_1 \neq t_3 - t_2 $.
The difference between two consecutive terms is not constant.
Final Answer: The given sequence is not an A.P.

v. $ 0, -4, -8, -12, \dots $

Given Sequence: $ 0, -4, -8, -12, \dots $
Here: $ t_1 = 0 $, $ t_2 = -4 $, $ t_3 = -8 $, $ t_4 = -12 $
Calculation:
$ t_2 - t_1 = -4 - 0 = -4 $
$ t_3 - t_2 = -8 - (-4) = -8 + 4 = -4 $
$ t_4 - t_3 = -12 - (-8) = -12 + 8 = -4 $
Since $ t_2 - t_1 = t_3 - t_2 = \dots = -4 = d $ (constant).
The difference between two consecutive terms is constant.
Final Answer: The given sequence is an A.P. and the common difference $ (d) = -4 $.

vi. $ -\frac{1}{5}, -\frac{1}{5}, -\frac{1}{5}, \dots $

Given Sequence: $ -\frac{1}{5}, -\frac{1}{5}, -\frac{1}{5}, \dots $
Here: $ t_1 = -\frac{1}{5} $, $ t_2 = -\frac{1}{5} $, $ t_3 = -\frac{1}{5} $
Calculation:
$ t_2 - t_1 = -\frac{1}{5} - \left(-\frac{1}{5}\right) = -\frac{1}{5} + \frac{1}{5} = 0 $
$ t_3 - t_2 = -\frac{1}{5} - \left(-\frac{1}{5}\right) = -\frac{1}{5} + \frac{1}{5} = 0 $
Since $ t_2 - t_1 = t_3 - t_2 = \dots = 0 = d $ (constant).
The difference between two consecutive terms is constant.
Final Answer: The given sequence is an A.P. and the common difference $ (d) = 0 $.

vii. $ 3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots $

Given Sequence: $ 3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots $
Here: $ t_1 = 3 $, $ t_2 = 3+\sqrt{2} $, $ t_3 = 3+2\sqrt{2} $, $ t_4 = 3+3\sqrt{2} $
Calculation:
$ t_2 - t_1 = (3+\sqrt{2}) - 3 = \sqrt{2} $
$ t_3 - t_2 = (3+2\sqrt{2}) - (3+\sqrt{2}) = \sqrt{2} $
$ t_4 - t_3 = (3+3\sqrt{2}) - (3+2\sqrt{2}) = \sqrt{2} $
Since $ t_2 - t_1 = t_3 - t_2 = \dots = \sqrt{2} = d $ (constant).
The difference between two consecutive terms is constant.
Final Answer: The given sequence is an A.P. and the common difference $ (d) = \sqrt{2} $.

viii. $ 127, 132, 137, \dots $

Given Sequence: $ 127, 132, 137, \dots $
Here: $ t_1 = 127 $, $ t_2 = 132 $, $ t_3 = 137 $
Calculation:
$ t_2 - t_1 = 132 - 127 = 5 $
$ t_3 - t_2 = 137 - 132 = 5 $
Since $ t_2 - t_1 = t_3 - t_2 = \dots = 5 = d $ (constant).
The difference between two consecutive terms is constant.
Final Answer: The given sequence is an A.P. and the common difference $ (d) = 5 $.

Practice Set 3.1: Question 2

Write an A.P. whose first term is $ a $ and common difference is $ d $ in each of the following.

Key Formula: The terms of an A.P. can be found using the sequence: $ t_1 = a $, $ t_2 = t_1 + d $, $ t_3 = t_2 + d $, $ \dots $

i. $ a = 10 $, $ d = 5 $

Given: $ a = 10 $, $ d = 5 $
Calculation:
$ t_1 = a = 10 $
$ t_2 = t_1 + d = 10 + 5 = 15 $
$ t_3 = t_2 + d = 15 + 5 = 20 $
$ t_4 = t_3 + d = 20 + 5 = 25 $
Final Answer: The required A.P. is 10, 15, 20, 25, ...

ii. $ a = -3 $, $ d = 0 $

Given: $ a = -3 $, $ d = 0 $
Calculation:
$ t_1 = a = -3 $
$ t_2 = t_1 + d = -3 + 0 = -3 $
$ t_3 = t_2 + d = -3 + 0 = -3 $
$ t_4 = t_3 + d = -3 + 0 = -3 $
Final Answer: The required A.P. is -3, -3, -3, -3, ...

iii. $ a = -7 $, $ d = \frac{1}{2} $

Given: $ a = -7 $, $ d = \frac{1}{2} $ (or $ 0.5 $)
Calculation:
$ t_1 = a = -7 $
$ t_2 = t_1 + d = -7 + \frac{1}{2} = \frac{-14 + 1}{2} = \frac{-13}{2} = -6.5 $
$ t_3 = t_2 + d = -6.5 + 0.5 = -6 $
$ t_4 = t_3 + d = -6 + \frac{1}{2} = \frac{-12 + 1}{2} = \frac{-11}{2} = -5.5 $
Final Answer: The required A.P. is -7, -6.5, -6, -5.5, ...

iv. $ a = -1.25 $, $ d = 3 $

Given: $ a = -1.25 $, $ d = 3 $
Calculation:
$ t_1 = a = -1.25 $
$ t_2 = t_1 + d = -1.25 + 3 = 1.75 $
$ t_3 = t_2 + d = 1.75 + 3 = 4.75 $
$ t_4 = t_3 + d = 4.75 + 3 = 7.75 $
Final Answer: The required A.P. is -1.25, 1.75, 4.75, 7.75, ...

v. $ a = 6 $, $ d = -3 $

Given: $ a = 6 $, $ d = -3 $
Calculation:
$ t_1 = a = 6 $
$ t_2 = t_1 + d = 6 + (-3) = 3 $
$ t_3 = t_2 + d = 3 + (-3) = 0 $
$ t_4 = t_3 + d = 0 + (-3) = -3 $
Final Answer: The required A.P. is 6, 3, 0, -3, ...

vi. $ a = -19 $, $ d = -4 $

Given: $ a = -19 $, $ d = -4 $
Calculation:
$ t_1 = a = -19 $
$ t_2 = t_1 + d = -19 + (-4) = -23 $
$ t_3 = t_2 + d = -23 + (-4) = -27 $
$ t_4 = t_3 + d = -27 + (-4) = -31 $
Final Answer: The required A.P. is -19, -23, -27, -31, ...

Practice Set 3.1: Question 3

Find the first term and common difference for each of the A.P.

i. $ 5, 1, -3, -7, \dots $

Given A.P.: $ 5, 1, -3, -7, \dots $
Here: $ t_1 = 5 $, $ t_2 = 1 $
Calculation:
First term $ a = t_1 = 5 $
Common difference $ d = t_2 - t_1 = 1 - 5 = -4 $
Final Answer: First term $ (a) = 5 $, common difference $ (d) = -4 $.

ii. $ 0.6, 0.9, 1.2, 1.5, \dots $

Given A.P.: $ 0.6, 0.9, 1.2, 1.5, \dots $
Here: $ t_1 = 0.6 $, $ t_2 = 0.9 $
Calculation:
First term $ a = t_1 = 0.6 $
Common difference $ d = t_2 - t_1 = 0.9 - 0.6 = 0.3 $
Final Answer: First term $ (a) = 0.6 $, common difference $ (d) = 0.3 $.

iii. $ 127, 135, 143, 151, \dots $

Given A.P.: $ 127, 135, 143, 151, \dots $
Here: $ t_1 = 127 $, $ t_2 = 135 $
Calculation:
First term $ a = t_1 = 127 $
Common difference $ d = t_2 - t_1 = 135 - 127 = 8 $
Final Answer: First term $ (a) = 127 $, common difference $ (d) = 8 $.

iv. $ \frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \dots $

Given A.P.: $ \frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \dots $
Here: $ t_1 = \frac{1}{4} $, $ t_2 = \frac{3}{4} $
Calculation:
First term $ a = t_1 = \frac{1}{4} $
Common difference $ d = t_2 - t_1 = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} $
Final Answer: First term $ (a) = \frac{1}{4} $, common difference $ (d) = \frac{1}{2} $.

Additional Textbook Questions (Page 55 & 56)

Question 1: Complete the given pattern

i. Circle Pattern

Pattern Number	1	2	3	4	5	6	7	8	9
Number of circles	1	3	5	7	9	11	13	15	17

Rule: Every pattern is formed by adding a circle in horizontal and vertical rows to the preceding pattern. The sequence is 1, 3, 5, 7, 9, 11, 13, 15, 17, ...

ii. Triangle Pattern

Pattern Number	1	2	3	4	5	6	7
Number of triangles	5	8	11	14	17	20	23

Rule: Every pattern is formed by adding 2 triangles horizontally and 1 triangle vertically to the preceding pattern. The sequence is 5, 8, 11, 14, 17, 20, 23, ...

Question 2: Show the positions of the terms by $ t_1, t_2, t_3, \dots $

i. $ 7, 7, 7, 7, \dots $
Here $ t_1 = 7 $, $ t_2 = 7 $, $ t_3 = 7 $, $ \dots $
ii. $ -2, -6, -10, -14, \dots $
Here $ t_1 = -2 $, $ t_2 = -6 $, $ t_3 = -10 $, $ \dots $

Question 3: Check whether there is any rule among the terms

i. $ 1, 4, 7, 10, 13, \dots $
Rule: By adding 3 to each term, we get the next term.
ii. $ 6, 12, 18, 24, \dots $
Rule: By adding 6 to each term, we get the next term.
iii. $ 3, 3, 3, 3, \dots $
Rule: By adding 0 to each term, we get the next term.
iv. $ 4, 16, 64, \dots $
Rule: By multiplying each term by 4, we get the next term. (Note: This is a Geometric Progression).
v. $ -1, -1.5, -2, -2.5, \dots $
Rule: By adding -0.5 to each term, we get the next term.
vi. $ 1^3, 2^3, 3^3, 4^3, \dots $
Rule: (Next term) = $ (\text{previous term's base} + 1)^3 $.

Conclusion: The similarity in sequences i, ii, iii, and v is that the next term is obtained by adding a particular fixed number to the previous term (Arithmetic Progression).

Question 4: Write one example of finite and infinite A.P. each.

Finite A.P.: Even natural numbers from 4 to 50:
4, 6, 8, \dots, 50
Infinite A.P.: Positive multiples of 5:
5, 10, 15, \dots

50 Arithmetic Progression (AP) Problems with Solutions & Formulas

Essential Arithmetic Progression Formulas

Key Formulas

General Term ($n$th term):
$$a_n = a + (n-1)d$$

Sum of $n$ terms:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$

Sum (using last term $l$):
$$S_n = \frac{n}{2}(a + l)$$

Common Difference:
$$d = a_n - a_{n-1}$$

Arithmetic Mean:
$$A = \frac{a+b}{2}$$

Sum of first $n$ natural numbers:
$$S_n = \frac{n(n+1)}{2}$$

Where $a = \text{first term}$, $d = \text{common difference}$, $n = \text{number of terms}$, $l = \text{last term}$.

Part 1: Basic Terms and Common Difference

1 Find the 10th term of the AP: 2, 7, 12, 17...

Solution Here, $a = 2$, $d = 7 - 2 = 5$, $n = 10$.
$$a_{10} = a + (10-1)d$$ $$a_{10} = 2 + 9(5) = 2 + 45 = 47$$ Answer: 47

2 Find the common difference of the AP: $\frac{1}{2}, \frac{3}{2}, \frac{5}{2}...$

Solution $d = a_2 - a_1 = \frac{3}{2} - \frac{1}{2} = \frac{2}{2} = 1$.
Answer: 1

3 Find the 15th term of the AP: 10, 7, 4...

Solution $a = 10$, $d = 7 - 10 = -3$, $n = 15$.
$a_{15} = 10 + 14(-3) = 10 - 42 = -32$.
Answer: -32

4 If $a = -5$ and $d = 2$, find the first 4 terms.

Solution Terms are $a, a+d, a+2d, a+3d$.
$-5, (-5+2), (-5+4), (-5+6)$
Answer: -5, -3, -1, 1

5 Find the $n$th term of the AP: 13, 8, 3, -2...

Solution $a = 13, d = -5$.
$a_n = 13 + (n-1)(-5) = 13 - 5n + 5 = 18 - 5n$.
Answer: $18 - 5n$

6 Which term of the AP 21, 18, 15... is -81?

Solution $a=21, d=-3, a_n = -81$.
$-81 = 21 + (n-1)(-3) \Rightarrow -102 = (n-1)(-3) \Rightarrow 34 = n-1 \Rightarrow n = 35$.
Answer: 35th term

7 Check if 301 is a term of 5, 11, 17, 23...

Solution $a=5, d=6$. Let $a_n = 301$.
$301 = 5 + (n-1)6 \Rightarrow 296 = 6(n-1) \Rightarrow n-1 = 49.33$.
Since $n$ is not an integer, 301 is not a term.
Answer: No

8 Find $k$ if $2k$, $k+10$, and $3k+2$ are in AP.

Solution If $x, y, z$ are in AP, $2y = x + z$.
$2(k+10) = 2k + (3k+2) \Rightarrow 2k + 20 = 5k + 2 \Rightarrow 18 = 3k \Rightarrow k = 6$.
Answer: $k=6$

9 How many two-digit numbers are divisible by 3?

Solution AP: 12, 15, ..., 99. $a=12, d=3, a_n=99$.
$99 = 12 + (n-1)3 \Rightarrow 87 = 3(n-1) \Rightarrow 29 = n-1 \Rightarrow n=30$.
Answer: 30

10 Find the 11th term from the end of the AP: 10, 7, 4... -62.

Solution Reverse the AP: $a = -62, d = +3$ (opposite of original $d=-3$).
$a_{11} = -62 + 10(3) = -62 + 30 = -32$.
Answer: -32

Part 2: Sum of Terms ($S_n$)

11 Find the sum of the first 20 terms of 1, 4, 7, 10...

Solution $a=1, d=3, n=20$.
$S_{20} = \frac{20}{2}[2(1) + 19(3)] = 10[2 + 57] = 10(59) = 590$.
Answer: 590

12 Find the sum of the first 100 positive integers.

Solution $S_n = \frac{n(n+1)}{2}$.
$S_{100} = \frac{100(101)}{2} = 50(101) = 5050$.
Answer: 5050

13 Find the sum of 2 + 4 + 6 + ... + 200.

Solution $a=2, l=200, n=100$.
$S_{100} = \frac{100}{2}(2 + 200) = 50(202) = 10100$.
Answer: 10100

14 Find the sum of the first 15 multiples of 8.

Solution $a=8, d=8, n=15$.
$S_{15} = \frac{15}{2}[2(8) + 14(8)] = \frac{15}{2}[16 + 112] = \frac{15}{2}(128) = 15(64) = 960$.
Answer: 960

15 How many terms of the AP 24, 21, 18... must be taken so that their sum is 78?

Solution $a=24, d=-3, S_n=78$.
$78 = \frac{n}{2}[48 + (n-1)(-3)] \Rightarrow 156 = n(48 - 3n + 3) \Rightarrow 156 = 51n - 3n^2$.
$n^2 - 17n + 52 = 0 \Rightarrow (n-4)(n-13)=0$.
Answer: $n=4$ or $n=13$

16 Find the sum of all odd numbers between 0 and 50.

Solution AP: 1, 3, 5... 49. Here $n = 25$.
$S_{25} = \frac{25}{2}(1 + 49) = \frac{25}{2}(50) = 25 \times 25 = 625$.
Answer: 625

17 The first and last terms of an AP are 1 and 11. If the sum is 36, find $n$.

Solution $S_n = \frac{n}{2}(a+l) \Rightarrow 36 = \frac{n}{2}(1+11) \Rightarrow 36 = 6n \Rightarrow n=6$.
Answer: 6

18 Find the sum of first 22 terms of an AP in which $d=7$ and 22nd term is 149.

Solution $a_{22} = a + 21d \Rightarrow 149 = a + 21(7) \Rightarrow a = 149 - 147 = 2$.
$S_{22} = \frac{22}{2}(2 + 149) = 11(151) = 1661$.
Answer: 1661

19 If $S_n = 3n^2 + n$, find the AP.

Solution $a_1 = S_1 = 3(1)^2 + 1 = 4$.
$S_2 = 3(2)^2 + 2 = 14$. $a_2 = S_2 - S_1 = 14 - 4 = 10$.
$d = 10 - 4 = 6$.
Answer: 4, 10, 16, 22...

20 Find the sum of first 15 terms of an AP whose $n$th term is $a_n = 9 - 5n$.

Solution $a_1 = 4, a_{15} = 9 - 75 = -66$.
$S_{15} = \frac{15}{2}(4 + (-66)) = \frac{15}{2}(-62) = 15(-31) = -465$.
Answer: -465

Part 3: Finding Unknowns (a, d, n)

21 Determine the AP whose 3rd term is 5 and the 7th term is 9.

Solution $a+2d=5$ and $a+6d=9$.
Subtracting: $4d=4 \Rightarrow d=1$. Substituting $d=1$: $a+2=5 \Rightarrow a=3$.
Answer: 3, 4, 5, 6...

22 Which term of the AP 3, 15, 27, 39... will be 132 more than its 54th term?

Solution $d=12$. Let $a_n = a_{54} + 132$.
$a + (n-1)d = a + 53d + 132$.
$(n-1)12 = 53(12) + 132 \Rightarrow (n-1)12 = 636 + 132 = 768$.
$n-1 = 64 \Rightarrow n = 65$.
Answer: 65th term

23 Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution Difference between corresponding terms of two APs with same $d$ is constant ($a_1 - a_2$).
Thus, difference is always 100.
Answer: 100

24 The sum of 4th and 8th terms of an AP is 24 and sum of 6th and 10th terms is 44. Find the AP.

Solution $(a+3d)+(a+7d)=24 \Rightarrow 2a+10d=24 \Rightarrow a+5d=12$.
$(a+5d)+(a+9d)=44 \Rightarrow 2a+14d=44 \Rightarrow a+7d=22$.
Solving: $2d=10 \Rightarrow d=5, a=-13$.
Answer: -13, -8, -3...

25 If 7 times the 7th term is equal to 11 times the 11th term, find the 18th term.

Solution $7(a+6d) = 11(a+10d) \Rightarrow 7a+42d = 11a+110d \Rightarrow 4a = -68d \Rightarrow a = -17d$.
$a_{18} = a + 17d = -17d + 17d = 0$.
Answer: 0

26 Find the number of terms in AP: 18, 15.5, 13, ..., -47.

Solution $a=18, d=-2.5, a_n=-47$.
$-47 = 18 + (n-1)(-2.5) \Rightarrow -65 = (n-1)(-2.5) \Rightarrow n-1 = 26 \Rightarrow n=27$.
Answer: 27

27 Subba Rao started work at an annual salary of 5000 with an increment of 200 each year. In which year did his income reach 7000?

Solution $a=5000, d=200, a_n=7000$.
$7000 = 5000 + (n-1)200 \Rightarrow 2000 = 200(n-1) \Rightarrow 10 = n-1 \Rightarrow n=11$.
Answer: 11th year

28 The sum of the first $n$ terms is $4n - n^2$. What is the first term? What is the 2nd term?

Solution $S_1 = 4(1) - 1 = 3$ (First term).
$S_2 = 4(2) - 4 = 4$. $a_2 = S_2 - S_1 = 4 - 3 = 1$.
Answer: 1st term = 3, 2nd term = 1

29 Find the sum of all 3-digit numbers divisible by 7.

Solution First: 105, Last: 994. $n = \frac{994-105}{7} + 1 = 128$.
$S = \frac{128}{2}(105 + 994) = 64(1099) = 70336$.
Answer: 70336

30 A sum of 700 is to be used to give 7 cash prizes. If each prize is 20 less than its preceding prize, find the value of each prize.

Solution $n=7, S_7=700, d=-20$.
$700 = \frac{7}{2}[2a + 6(-20)] \Rightarrow 200 = 2a - 120 \Rightarrow 320 = 2a \Rightarrow a=160$.
Answer: 160, 140, 120, 100, 80, 60, 40

Part 4: Word Problems & Higher Order Thinking

31 Ramkali saved 5 in the first week of a year and then increased her weekly savings by 1.75. If in the $n$th week her savings become 20.75, find $n$.

Solution $a=5, d=1.75, a_n=20.75$.
$20.75 = 5 + (n-1)1.75 \Rightarrow 15.75 = (n-1)1.75 \Rightarrow n-1 = 9 \Rightarrow n=10$.
Answer: 10

32 The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first 16 terms of the AP.

Solution $a_3 + a_7 = 6 \Rightarrow 2a + 8d = 6 \Rightarrow a + 4d = 3 \Rightarrow a = 3-4d$.
$a_3 \times a_7 = 8 \Rightarrow (a+2d)(a+6d) = 8$.
Substitute $a$: $(3-2d)(3+2d) = 8 \Rightarrow 9 - 4d^2 = 8 \Rightarrow d^2 = 1/4 \Rightarrow d = \pm 1/2$.
If $d=1/2, a=1, S_{16} = 76$. If $d=-1/2, a=5, S_{16} = 20$.
Answer: 76 or 20

33 Find the sum of odd numbers between 0 and 50.

Solution Repeated Concept: $n=25$, Sum = $n^2 = 25^2 = 625$.
Answer: 625

34 A manufacturer of TV sets produced 600 sets in the 3rd year and 700 sets in the 7th year. Assuming uniform increase, find production in 1st year.

Solution $a+2d=600$ and $a+6d=700$.
$4d=100 \Rightarrow d=25$. $a = 600 - 50 = 550$.
Answer: 550 sets

35 In Q34, what is the total production in 7 years?

Solution $S_7 = \frac{7}{2}(550 + 700) = \frac{7}{2}(1250) = 7(625) = 4375$.
Answer: 4375 sets

36 Which term of the AP 121, 117, 113... is its first negative term?

Solution $a=121, d=-4$. We need $a_n < 0$.
$121 + (n-1)(-4) < 0 \Rightarrow 121 - 4n + 4 < 0 \Rightarrow 125 < 4n \Rightarrow n > 31.25$.
Answer: 32nd term

37 The houses of a row are numbered consecutively from 1 to 49. Is there a value of $x$ such that sum of numbers preceding $x$ is equal to sum of numbers following it?

Solution $S_{x-1} = S_{49} - S_x$.
$\frac{x-1}{2}(1 + x-1) = \frac{49}{2}(50) - \frac{x}{2}(1+x)$.
$x(x-1) = 2450 - x(x+1) \Rightarrow x^2 - x = 2450 - x^2 - x \Rightarrow 2x^2 = 2450 \Rightarrow x^2=1225$.
$x = \sqrt{1225} = 35$.
Answer: Yes, x = 35

38 Find the middle term of the AP 6, 13, 20... 216.

Solution $a=6, d=7$. $216 = 6 + (n-1)7 \Rightarrow 210 = 7(n-1) \Rightarrow n=31$.
Middle term is $\frac{31+1}{2} = 16$th term.
$a_{16} = 6 + 15(7) = 6 + 105 = 111$.
Answer: 111

39 If the $p$th term is $q$ and $q$th term is $p$, prove that $n$th term is $(p+q-n)$.

Solution $a+(p-1)d=q$ and $a+(q-1)d=p$.
Subtract: $(p-q)d = q-p \Rightarrow d = -1$.
Substitute: $a - p + 1 = q \Rightarrow a = p+q-1$.
$a_n = (p+q-1) + (n-1)(-1) = p+q-1-n+1 = p+q-n$. (Proved)

40 Split 207 into 3 parts such that these are in AP and the product of the two smaller parts is 4623.

Solution Let parts be $a-d, a, a+d$. Sum $= 3a = 207 \Rightarrow a=69$.
Product: $(69-d)(69) = 4623 \Rightarrow 69-d = 67 \Rightarrow d=2$.
Parts: $67, 69, 71$.
Answer: 67, 69, 71

Part 5: Rapid Fire & Properties

41 Is the sequence $a_n = 2n^2 + 1$ an AP?

Solution $a_1=3, a_2=9, a_3=19$. Differences: $6, 10$. Not constant.
Answer: No

42 If $a, b, c$ are in AP, then $2b = \dots$?

Solution $2b = a + c$ (Arithmetic Mean property).
Answer: $a+c$

43 Find the sum of first $n$ odd natural numbers.

Solution Formula for sum of $n$ odd numbers is $n^2$.
Answer: $n^2$

44 If the common difference is 5, what is $a_{18} - a_{13}$?

Solution $a_{18} - a_{13} = (a+17d) - (a+12d) = 5d$.
$5(5) = 25$.
Answer: 25

45 Find $a$ if $a, a-2, 3a$ are in AP.

Solution $2(a-2) = a + 3a \Rightarrow 2a - 4 = 4a \Rightarrow -4 = 2a \Rightarrow a = -2$.
Answer: -2

46 If the sum of first $p$ terms is equal to the sum of first $q$ terms, find $S_{p+q}$.

Solution Standard property: If $S_p = S_q$, then $S_{p+q} = 0$.
Answer: 0

47 Find the sum of even numbers between 5 and 15.

Solution Numbers: 6, 8, 10, 12, 14. Sum $= 50$.
Answer: 50

48 In an AP, $a_m = n$ and $a_n = m$. Find $a_{m+n}$.

Solution Standard result: $d = -1$, $a = m+n-1$.
$a_{m+n} = a + (m+n-1)d = (m+n-1) - (m+n-1) = 0$.
Answer: 0

49 Three numbers are in AP and their sum is 24. Find the middle term.

Solution Let terms be $a-d, a, a+d$. Sum $= 3a = 24 \Rightarrow a=8$.
Answer: 8

50 If $18, a, b, -3$ are in AP, find $a+b$.

Solution Sum of equidistant terms is constant. $a+b = 18 + (-3) = 15$.
Answer: 15

46 Arithmetic Progression Questions for Reference.

46 Arithmetic Progression Questions with Solutions

A complete question bank for Class 10 & 12 Board Prep provided by Omtex Classes.

Part 1: Basic Concepts and Nth Term

Question 1

Find the common difference of the AP: $ 5, 8, 11, 14, \dots $

Solution: Common difference $ d = a_2 - a_1 $
$ d = 8 - 5 = 3 $.
Answer: 3

Question 2

Find the 10th term of the AP: $ 2, 7, 12, \dots $

Solution: Here, $ a = 2 $, $ d = 7 - 2 = 5 $, $ n = 10 $.
Using $ a_n = a + (n-1)d $:
$ a_{10} = 2 + (10-1)5 = 2 + 9(5) = 2 + 45 = 47 $.
Answer: 47

Question 3

Check if the list of numbers $ 2, 4, 8, 16, \dots $ forms an AP.

Solution: $ a_2 - a_1 = 4 - 2 = 2 $
$ a_3 - a_2 = 8 - 4 = 4 $
Since the difference is not constant, it is not an AP.

Question 4

Find the first term $ a $ if the common difference is -3 and the 10th term is -25.

Solution: Given $ d = -3, a_{10} = -25 $.
$ a_{10} = a + 9d $
$ -25 = a + 9(-3) $
$ -25 = a - 27 \Rightarrow a = 2 $.
Answer: 2

Question 5

Find the general term ($ n $th term) of the AP: $ 13, 8, 3, -2, \dots $

Solution: $ a = 13, d = 8 - 13 = -5 $.
$ a_n = 13 + (n-1)(-5) = 13 - 5n + 5 $.
Answer: $ a_n = 18 - 5n $

Question 6

Which term of the AP $ 3, 8, 13, 18, \dots $ is 78?

Solution: $ a=3, d=5, a_n=78 $.
$ 78 = 3 + (n-1)5 $
$ 75 = 5(n-1) \Rightarrow 15 = n - 1 \Rightarrow n = 16 $.
Answer: 16th term

Question 7

Find the 20th term from the last term of the AP: $ 3, 8, 13, \dots, 253 $.

Solution: Reverse the AP: $ 253, \dots, 13, 8, 3 $.
New $ a = 253 $, New $ d = -5 $.
$ a_{20} = 253 + (19)(-5) = 253 - 95 = 158 $.
Answer: 158

Question 8

Determine the AP whose 3rd term is 5 and the 7th term is 9.

Solution: $ a+2d = 5 $ (i)
$ a+6d = 9 $ (ii)
Subtract (i) from (ii): $ 4d = 4 \Rightarrow d = 1 $.
Substitute in (i): $ a + 2(1) = 5 \Rightarrow a = 3 $.
Answer: 3, 4, 5, 6, ...

Question 9

How many two-digit numbers are divisible by 3?

Solution: AP: $ 12, 15, \dots, 99 $.
$ a=12, d=3, a_n=99 $.
$ 99 = 12 + (n-1)3 $
$ 87 = 3(n-1) \Rightarrow 29 = n - 1 \Rightarrow n = 30 $.
Answer: 30

Question 10

Find the middle term of the AP: $ 6, 13, 20, \dots, 216 $.

Solution: $ a=6, d=7, a_n=216 $.
$ 216 = 6 + (n-1)7 \Rightarrow 210 = 7(n-1) \Rightarrow n=31 $.
Middle term is $ \frac{31+1}{2} = 16 $th term.
$ a_{16} = 6 + 15(7) = 6 + 105 = 111 $.
Answer: 111

Part 2: Finding Unknowns

Question 11

For what value of $ k $ are $ 2k, k+10, $ and $ 3k+2 $ in AP?

Solution: If $ a, b, c $ are in AP, $ 2b = a + c $.
$ 2(k+10) = 2k + (3k+2) $
$ 2k + 20 = 5k + 2 $
$ 18 = 3k \Rightarrow k = 6 $.
Answer: 6

Question 12

If $ x+2, 2x, 2x+3 $ are in AP, find $ x $.

Solution: $ 2(2x) = (x+2) + (2x+3) $
$ 4x = 3x + 5 \Rightarrow x = 5 $.
Answer: 5

Question 13

Find $ a, b, c $ such that the numbers $ a, 7, b, 23, c $ are in AP.

Solution: $ a_2 = 7 \Rightarrow a+d=7 $. $ a_4 = 23 \Rightarrow a+3d=23 $.
Subtracting: $ 2d = 16 \Rightarrow d = 8 $.
$ a = -1, b = 15, c = 31 $.
Answer: a=-1, b=15, c=31

Question 14

Is 301 a term of the AP $ 5, 11, 17, 23, \dots $?

Solution: $ a=5, d=6 $.
$ 301 = 5 + (n-1)6 \Rightarrow 296 = 6(n-1) $.
$ n-1 = 296/6 = 49.33 $.
Since $ n $ is not an integer, 301 is not a term.

Question 15

Find the value of $ p $ if the numbers $ 2p-1, 3p+1, 11 $ are in AP.

Solution: $ 2(3p+1) = (2p-1) + 11 $
$ 6p + 2 = 2p + 10 $
$ 4p = 8 \Rightarrow p = 2 $.
Answer: 2

Part 3: Sum of n Terms (Sn)

Question 16

Find the sum of the first 20 terms of the AP: $ 1, 4, 7, 10, \dots $

Solution: $ a=1, d=3, n=20 $.
$ S_n = \frac{n}{2}[2a+(n-1)d] $
$ S_{20} = \frac{20}{2}[2(1) + 19(3)] = 10[2 + 57] = 10(59) = 590 $.
Answer: 590

Question 17

Find the sum of the first 100 positive integers.

Solution: $ S_n = \frac{n(n+1)}{2} $
$ S_{100} = \frac{100 \times 101}{2} = 50 \times 101 = 5050 $.
Answer: 5050

Question 18

Find the sum of: $ 34 + 32 + 30 + \dots + 10 $.

Solution: $ a=34, d=-2, l=10 $.
Find $ n $: $ 10 = 34 + (n-1)(-2) \Rightarrow -24 = -2(n-1) \Rightarrow n=13 $.
$ S_{13} = \frac{13}{2}(a+l) = \frac{13}{2}(34+10) = \frac{13}{2}(44) = 13 \times 22 = 286 $.
Answer: 286

Question 19

The sum of the first $ n $ terms of an AP is given by $ S_n = 3n^2 + n $. Find the 2nd term.

Solution: $ a_1 = S_1 = 3(1)^2 + 1 = 4 $.
$ S_2 = 3(2)^2 + 2 = 14 $.
$ a_2 = S_2 - S_1 = 14 - 4 = 10 $.
Answer: 10

Question 20

How many terms of the AP $ 24, 21, 18, \dots $ must be taken so that their sum is 78?

Solution: $ a=24, d=-3, S_n=78 $.
$ 78 = \frac{n}{2}[48 + (n-1)(-3)] \Rightarrow 156 = n[48 - 3n + 3] $
$ 156 = 51n - 3n^2 \Rightarrow 3n^2 - 51n + 156 = 0 \Rightarrow n^2 - 17n + 52 = 0 $.
$ (n-4)(n-13) = 0 $. Both are valid.
Answer: 4 or 13

Question 21

Find the sum of the first 15 multiples of 8.

Solution: AP: $ 8, 16, \dots $. $ a=8, d=8, n=15 $.
$ S_{15} = \frac{15}{2}[2(8) + 14(8)] = \frac{15}{2}(16 + 112) = \frac{15}{2}(128) = 15 \times 64 = 960 $.
Answer: 960

Question 22

Find the sum of all odd numbers between 0 and 50.

Solution: AP: $ 1, 3, 5, \dots, 49 $. $ a=1, d=2, l=49 $.
$ 49 = 1 + (n-1)2 \Rightarrow 48 = 2(n-1) \Rightarrow n=25 $.
$ S_{25} = \frac{25}{2}(1+49) = \frac{25}{2}(50) = 625 $.
Answer: 625

Question 23

If the sum of first 7 terms is 49 and that of 17 terms is 289, find the sum of first $ n $ terms.

Solution: Given pattern: $ S_n = n^2 $.
(Proof: $ S_7=7^2, S_{17}=17^2 $).
Answer: $ n^2 $

Question 24

Find the sum of the first 40 positive integers divisible by 6.

Solution: $ a=6, d=6, n=40 $.
$ S_{40} = \frac{40}{2}[12 + 39(6)] = 20(12 + 234) = 20(246) = 4920 $.
Answer: 4920

Question 25

The first and the last terms of an AP are 17 and 350. If the common difference is 9, how many terms are there and what is their sum?

Solution: $ a=17, l=350, d=9 $.
$ 350 = 17 + (n-1)9 \Rightarrow 333 = 9(n-1) \Rightarrow 37 = n-1 \Rightarrow n=38 $.
$ S_{38} = \frac{38}{2}(17+350) = 19(367) = 6973 $.
Answer: n=38, Sum=6973

Part 4: Word Problems

Question 26

Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

Solution: AP: $ 5000, 5200, \dots, 7000 $.
$ a=5000, d=200, a_n=7000 $.
$ 7000 = 5000 + (n-1)200 \Rightarrow 2000 = 200(n-1) \Rightarrow 10 = n-1 \Rightarrow n=11 $.
Year = $ 1995 + 10 = 2005 $.
Answer: 2005

Question 27

A sum of Rs 700 is to be used to give seven cash prizes. If each prize is Rs 20 less than its preceding prize, find the value of each prize.

Solution: $ S_7 = 700, n=7, d=-20 $.
$ 700 = \frac{7}{2}[2a + 6(-20)] \Rightarrow 200 = 2a - 120 \Rightarrow 2a = 320 \Rightarrow a = 160 $.
Answer: 160, 140, 120, 100, 80, 60, 40

Question 28

In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted?

Solution: Trees per class: $ 3 \times 1, 3 \times 2, \dots, 3 \times 12 $.
AP: $ 3, 6, 9, \dots, 36 $. $ n=12 $.
$ S_{12} = \frac{12}{2}(3+36) = 6(39) = 234 $.
Answer: 234 trees

Question 29

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed?

Solution: $ S_n = 200, a=20, d=-1 $.
$ 200 = \frac{n}{2}[40 + (n-1)(-1)] \Rightarrow 400 = n(41-n) = 41n - n^2 $.
$ n^2 - 41n + 400 = 0 $. Factors of 400 summing to 41 are 16 and 25.
If $ n=25, a_{25} = 20 - 24 = -4 $ (impossible). So $ n=16 $.
Answer: 16 rows

Question 30

Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75. If in the $ n $th week, her weekly savings become Rs 20.75, find $ n $.

Solution: $ a=5, d=1.75, a_n=20.75 $.
$ 20.75 = 5 + (n-1)1.75 $
$ 15.75 = (n-1)1.75 \Rightarrow n-1 = 9 \Rightarrow n=10 $.
Answer: 10

Part 5: Advanced & Properties

Question 31

Find three numbers in AP whose sum is 24 and whose product is 440.

Solution: Let terms be $ a-d, a, a+d $.
Sum: $ 3a = 24 \Rightarrow a = 8 $.
Product: $ (8-d)(8)(8+d) = 440 $
$ 64 - d^2 = 55 \Rightarrow d^2 = 9 \Rightarrow d = \pm 3 $.
Terms: $ 5, 8, 11 $.
Answer: 5, 8, 11

Question 32

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms.

Solution: $ (a+3d) + (a+7d) = 24 \Rightarrow 2a + 10d = 24 \Rightarrow a+5d=12 $.
$ (a+5d) + (a+9d) = 44 \Rightarrow 2a + 14d = 44 \Rightarrow a+7d=22 $.
Solving: $ 2d=10 \Rightarrow d=5 $. $ a = -13 $.
Answer: -13, -8, -3

Question 33

Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623.

Solution: Let parts be $ a-d, a, a+d $. Sum $ 3a = 207 \Rightarrow a=69 $.
Smaller parts: $ (69-d)(69) = 4623 $.
$ 69-d = 67 \Rightarrow d=2 $.
Terms: $ 67, 69, 71 $.
Answer: 67, 69, 71

Question 34

Find the sum of all integers between 100 and 550 which are divisible by 9.

Solution: First term $ 108 $, Last term $ 549 $. $ d=9 $.
$ 549 = 108 + (n-1)9 \Rightarrow 441 = 9(n-1) \Rightarrow 49 = n-1 \Rightarrow n=50 $.
$ S_{50} = \frac{50}{2}(108+549) = 25(657) = 16425 $.
Answer: 16425

Question 35

If the $ m $th term of an AP is $ 1/n $ and the $ n $th term is $ 1/m $, show that the $ mn $th term is 1.

Solution: $ a + (m-1)d = 1/n $ (1)
$ a + (n-1)d = 1/m $ (2)
Subtracting: $ (m-n)d = \frac{m-n}{mn} \Rightarrow d = \frac{1}{mn} $.
Substituting $ d $: $ a = \frac{1}{mn} $.
$ a_{mn} = \frac{1}{mn} + (mn-1)\frac{1}{mn} = \frac{1 + mn - 1}{mn} = 1 $.
Answer: 1 (Proved)

Question 36

The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles.

Solution: Let angles be $ a-d, a, a+d $. Sum is 180.
$ 3a = 180 \Rightarrow a = 60^\circ $.
$ a+d = 2(a-d) \Rightarrow 60+d = 120 - 2d \Rightarrow 3d = 60 \Rightarrow d = 20 $.
Angles: $ 40^\circ, 60^\circ, 80^\circ $.
Answer: 40°, 60°, 80°

Question 37

Find the sum of the integers between 1 and 100 which are not divisible by 3.

Solution: Total sum (1 to 100) = 5050.
Sum of numbers divisible by 3 ($ 3, 6, \dots, 99 $): $ n=33 $.
$ S_3 = \frac{33}{2}(3+99) = \frac{33}{2}(102) = 33 \times 51 = 1683 $.
Required Sum = $ 5050 - 1683 = 3367 $.
Answer: 3367

Question 38

If 9 times the 9th term of an AP is equal to 13 times the 13th term, find the 22nd term.

Solution: $ 9(a+8d) = 13(a+12d) $
$ 9a + 72d = 13a + 156d $
$ -4a = 84d \Rightarrow a = -21d \Rightarrow a + 21d = 0 $.
Since $ a_{22} = a+21d $, answer is 0.
Answer: 0

Question 39

If the sum of first $ m $ terms of an AP is same as the sum of its first $ n $ terms ($ m \neq n $), find the sum of its first $ (m+n) $ terms.

Solution: $ S_m = S_n \Rightarrow \frac{m}{2}[2a+(m-1)d] = \frac{n}{2}[2a+(n-1)d] $.
Simplify: $ 2a(m-n) + d(m^2-m - n^2+n) = 0 $.
$ 2a(m-n) + d[(m-n)(m+n) - (m-n)] = 0 $.
Divide by $ m-n $: $ 2a + d(m+n-1) = 0 $.
$ S_{m+n} = \frac{m+n}{2}[2a + (m+n-1)d] = \frac{m+n}{2}(0) = 0 $.
Answer: 0

Question 40

The digits of a positive integer having three digits are in AP and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.

Solution: Digits $ a-d, a, a+d $. Sum $ 3a=15 \Rightarrow a=5 $.
Number: $ 100(5-d) + 10(5) + (5+d) $.
Reversed: $ 100(5+d) + 10(5) + (5-d) $.
Difference: $ -99(2d) = -594 \Rightarrow 198d = 594 \Rightarrow d = 3 $.
Digits: $ 8, 5, 2 $. Number: 852.
Answer: 852

Part 6: HOTS (Higher Order Thinking Skills)

Question 41

Determine $ k $ so that $ k^2 + 4k + 8, 2k^2 + 3k + 6, $ and $ 3k^2 + 4k + 4 $ are in AP.

Solution: $ 2b = a+c $.
$ 2(2k^2+3k+6) = (k^2+4k+8) + (3k^2+4k+4) $
$ 4k^2+6k+12 = 4k^2+8k+12 $
$ 6k = 8k \Rightarrow 2k = 0 \Rightarrow k = 0 $.
Answer: 0

Question 42

The sum of the first $ n $ terms of two APs are in the ratio $ (7n+1):(4n+27) $. Find the ratio of their $ m $th terms.

Solution: Ratio of sums $ \frac{S_n}{S'_n} = \frac{2a+(n-1)d}{2a'+(n-1)d'} = \frac{7n+1}{4n+27} $.
To find ratio of $ m $th term ($ \frac{a+(m-1)d}{a'+(m-1)d'} $), replace $ n $ with $ 2m-1 $.
$ \frac{a_m}{a'_m} = \frac{7(2m-1)+1}{4(2m-1)+27} = \frac{14m-6}{8m+23} $.
Answer: $ (14m-6):(8m+23) $

Question 43

Find the sum of all two-digit numbers which when divided by 4, yield 1 as remainder.

Solution: AP: $ 13, 17, \dots, 97 $.
$ 97 = 13 + (n-1)4 \Rightarrow 84 = 4(n-1) \Rightarrow 21 = n-1 \Rightarrow n=22 $.
$ S_{22} = \frac{22}{2}(13+97) = 11(110) = 1210 $.
Answer: 1210

Question 44

Solve the equation: $ 1 + 4 + 7 + 10 + \dots + x = 287 $.

Solution: $ a=1, d=3 $. Let total terms be $ n $.
$ 287 = \frac{n}{2}[2 + (n-1)3] \Rightarrow 574 = n(3n-1) $.
$ 3n^2 - n - 574 = 0 $.
Solving for $ n $: $ n = 14 $.
$ x = a_{14} = 1 + 13(3) = 1 + 39 = 40 $.
Answer: x = 40

Question 45

The 4th term of an AP is equal to 3 times the first term and the 7th term exceeds twice the 3rd term by 1. Find the first term and common difference.

Solution: $ a+3d = 3a \Rightarrow 3d = 2a $ (i)
$ a+6d = 2(a+2d) + 1 \Rightarrow a+6d = 2a+4d+1 \Rightarrow 2d = a+1 $.
From (i), $ a = 1.5d $. Substitute: $ 2d = 1.5d + 1 \Rightarrow 0.5d = 1 \Rightarrow d = 2 $.
$ a = 1.5(2) = 3 $.
Answer: a=3, d=2

Question 46

If $ S_1, S_2, S_3 $ are the sums of $ n $ terms of three APs, the first term of each being 1 and the respective common differences being 1, 2, 3, prove that $ S_1 + S_3 = 2S_2 $.

Solution: $ S_1 = \frac{n}{2}[2 + (n-1)1] = \frac{n(n+1)}{2} $.
$ S_2 = \frac{n}{2}[2 + (n-1)2] = n^2 $.
$ S_3 = \frac{n}{2}[2 + (n-1)3] = \frac{n(3n-1)}{2} $.
$ S_1 + S_3 = \frac{n}{2}(n+1 + 3n-1) = \frac{n}{2}(4n) = 2n^2 = 2S_2 $.
Answer: Proved

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Class 10 Maths Part 1 Chapter 3 Arithmetic Progression Problem Set 3 Solutions | Maharashtra Board

Question 1: Choose the correct alternative answer for each of the following sub questions.

i. The sequence -10, -6, -2, 2, ...

ii. First four terms of an A.P. are ..., whose first term is -2 and common difference is -2.

iii. What is the sum of the first 30 natural numbers?

iv. For a given A.P., \( t_{7} = 4 \), \( d = -4 \), then \( a = \)

v. For a given A.P., \( a = 3.5 \), \( d = 0 \), \( n = 101 \), then \( t_{n} = \)

vi. In an A.P., the first two terms are -3, 4, then the \( 21^{st} \) term is ....

vii. If for any A.P. \( d = 5 \), then \( t_{18} - t_{13} = \dots \)

viii. Sum of first five multiples of 3 is ...

ix. 15, 10, 5, ... In this A.P. sum of first 10 terms is...

x. In an A.P. \( 1^{st} \) term is 1 and the last term is 20. The sum of all terms is 399, then \( n = \)

Question 2

Solution:

Question 3

Solution:

Question 4

Solution:

Question 5

Solution:

Question 6

Solution:

Question 7

Solution:

Question 8

Solution:

Question 9

Solution:

Question 10

Solution:

Question 11

Solution:

Question 12

Solution:

Question 13

Solution:

Question 14

Solution & Completed Activity:

Maharashtra Board Class 10 Maths Part 1 Chapter 3 Arithmetic Progression Practice Set 3.4 Solutions

Practice Set 3.4 Questions and Answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Maharashtra Board Class 10 Maths Part 1 Practice Set 3.3 Solutions

Practice Set 3.3 Solutions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Additional Question (Textbook Pg. No. 71)

Class 10 Maths Part 1 Chapter 3 Arithmetic Progression Practice Set 3.2 Solutions (Maharashtra Board)

Question 1: Write the correct number in the given boxes from the following A.P.

Question 2: Decide whether following sequence is an A.P., if so find the 20th term of the progression.

Question 3: Given Arithmetic Progression is 12, 16, 20, 24, ... Find the 24th term of this progression.

Question 4: Find the 19th term of the following A.P. 7, 13, 19, 25, ...

Question 5: Find the 27th term of the following A.P. 9, 4, -1, -6, -11, ...

Question 6: Find how many three-digit natural numbers are divisible by 5.

Question 7: The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41st term of that A.P.

Question 8: 11, 8, 5, 2, ... In this A.P. which term is number -151?

Question 9: In the natural numbers from 10 to 250, how many are divisible by 4?

Question 10: In an A.P., the 17th term is 7 more than its 10th term. Find the common difference.

Application Question 1: Kabir's Height

Application Question 2: Testing Sequence Terms

Part i: Is it an A.P.?

Part ii: What will be the 100th term?

Part iii: Check whether 92 is in this A.P.

Part iv: Check whether 61 is in this A.P.

Class 10 Maths Part 1 Practice Set 3.1 Solutions | Maharashtra Board

Practice Set 3.1: Question 1

i. \( 2, 4, 6, 8, \dots \)

ii. \( 2, \frac{5}{2}, 3, \frac{7}{2}, \dots \)

iii. \( -10, -6, -2, 2, \dots \)