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.