Practice Set 3.3 Arithmetic Progression Class 10th Solutions

Practice Set 3.3 Arithmetic Progression

Class 10th Mathematics Part 1 MHB Solution

Question 1

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

Given: \( a = 6, d = 3, S_{27} = ? \)

Answer:

Given: First term \( a = 6 \)
Common Difference \( d = 3 \)
To find: \( S_{27} \) where \( n = 27 \)

By using sum of nth term of an A.P.:

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

Where,
\( n = \) no. of terms
\( a = \) first term
\( d = \) common difference
\( S_n = \) sum of n terms

Thus, Substituting given value in formula we can find the value of \( S_{27} \):

\( S_{27} = \frac{27}{2} [2 \times 6 + (27 - 1) \times 3] \)

\( S_{27} = \frac{27}{2} [12 + (26) \times 3] \)

\( S_{27} = \frac{27}{2} [12 + 78] \)

\( S_{27} = \frac{27}{2} \times 90 \)

\( S_{27} = 27 \times 45 = 1215 \)

Thus, \( S_{27} = 1215 \)

Question 2

Find the sum of first 123 even natural numbers.

Answer:

List of first 123 even natural number is: 2, 4, 6,.......

• Where first term \( a = 2 \)
• Second term \( t_1 = 4 \)
• Third term \( t_2 = 6 \)

Thus, common difference \( d = t_2 – t_1 = 6 – 4 = 2 \)
\( n = 123 \)

By using sum of nth term of an A.P.:

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

Where,
\( n = \) no. of terms
\( a = \) first term
\( d = \) common difference
\( S_n = \) sum of n terms

Thus, Substituting given value in formula we can find the value of \( S_n \):

\( S_{123} = \frac{123}{2} [2 \times 2 + (123 - 1) \times 2] \)

\( S_{123} = \frac{123}{2} [4 + 122 \times 2] \)

\( S_{123} = \frac{123}{2} [4 + 244] \)

\( S_{123} = \frac{123}{2} \times 248 \)

\( S_{123} = 123 \times 124 \)

Thus, \( S_n = 15252 \)

Question 3

Find the sum of all even numbers from 1 to 350.

Answer:

List of even natural number between 1 to 350 is: 2, 4, 6,.......348

• Where first term \( a = 2 \)
• Second term \( t_1 = 4 \)
• Third term \( t_2 = 6 \)

Thus, common difference \( d = t_2 – t_1 = 6 – 4 = 2 \)

\( t_n = 348 \) (As we have to find the sum of even numbers between 1 and 350 therefore excluding 350)

Now, By using nth term of an A.P. formula:

\( t_n = a + (n – 1)d \)

Where \( n = \) no. of terms, \( a = \) first term, \( d = \) common difference, \( t_n = \) nth terms.

We can find value of "n" by substituting all the value in formula we get,

\( 348 = 2 + (n – 1) \times 2 \)

\( 348 – 2 = 2(n – 1) \)

\( 346 = 2(n – 1) \)

\( \frac{346}{2} = n - 1 \)

\( 173 = n - 1 \)

\( n = 173 + 1 = 174 \)

Now, By using sum of nth term of an A.P. we will find its sum:

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

Substituting given value in formula we can find the value of \( S_n \):

\( S_{174} = \frac{174}{2} [2 \times 2 + (174 - 1) \times 2] \)

\( S_{174} = 87 [4 + 173 \times 2] \)

\( S_{174} = 87 [4 + 346] \)

\( S_{174} = 87 \times 350 \)

Thus, \( S_{174} = 30,450 \)

Question 4

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

Answer:

Given: \( t_{19} = 52 \) and \( t_{38} = 128 \)
To find: value of "a" and "d"

Using nth term of an A.P. formula:

\( t_n = a + (n – 1)d \)

Let, \( t_{19} = a + (19 – 1) d \)

\( 52 = a + 18 d .....(1) \)

\( t_{38} = a + (38 – 1) d \)

\( 128 = a + 37 d .....(2) \)

Subtracting eq. (1) from eq. (2), we get,

\( 128 – 52 = (a – a) + (37 d – 18 d) \)

\( 76 = 19 d \)

\( d = \frac{76}{19} = 4 \)

Substitute value of "d" in eq. (1) to get value of "a"

\( 52 = a + 18 \times 4 \)

\( 52 = a + 72 \)

\( a = 52 – 72 = – 20 \)

Now, to find value of \( S_{56} \) we will use formula of sum of n terms:

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

\( S_{56} = \frac{56}{2} [2(-20) + (56-1)4] \)

\( S_{56} = 28 \times [ – 40 + 55 \times 4] \)

\( S_{56} = 28 \times [ – 40 + 220] \)

\( S_{56} = 28 \times 180 = 5040 \)

Thus, \( S_{56} = 5040 \)

Question 5

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

Answer:

List of natural number divisible by 4 between 1 to 140 is: 4, 8, 12,.......136

• Where first term \( a = 4 \)
• Second term \( t_1 = 8 \)
• Third term \( t_2 = 12 \)

Thus, common difference \( d = t_2 – t_1 = 12 – 8 = 4 \)
\( t_n = 136 \)

Now, By using nth term of an A.P. formula:

\( t_n = a + (n – 1)d \)

We can find value of "n" by substituting all the value in formula we get,

\( 136 = 4 + (n – 1) \times 4 \)

\( 136 – 4 = 4(n – 1) \)

\( 132 = 4(n – 1) \)

\( n – 1 = 33 \)

\( n = 33 + 1 = 34 \)

Now, By using sum of nth term of an A.P. we will find its sum:

\( S_{34} = \frac{34}{2} [2a + (34-1)d] \)

Alternatively using direct substitution as per text:

\( S_{34} = 17 \times [2(4) + 33 \times 4] \)

\( S_{34} = 17 \times [8 + 132] \)

\( S_{34} = 17 \times 140 = 2380 \)

Thus, \( S_{34} = 2380 \)

Question 6

Sum of first 55 terms in an A.P. is 3300, find its 28th term.

Answer:

Given: \( S_{55} = 3300 \) where \( n = 55 \)

Now, By using sum of nth term of an A.P. we will find its sum:

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

Thus, on substituting the given value in formula we get,

\( 3300 = \frac{55}{2} [2a + (55 - 1)d] \)

\( 3300 = \frac{55}{2} [2a + 54d] \)

\( 3300 = 55 \times [a + 27d] \) (Taking 2 common and cancelling)

\( \frac{3300}{55} = a + 27d \)

\( a + 27d = 60 .....(1) \)

We need to find value of 28th term i.e \( t_{28} \)

Now, By using nth term of an A.P. formula:

\( t_n = a + (n – 1)d \)

We can find value of \( t_{28} \) by substituting all the value in formula we get,

\( t_{28} = a + (28 – 1) d \)

\( t_{28} = a + 27 d \)

From eq. (1) we get,

\( t_{28} = a + 27 d = 60 \)

\( t_{28} = 60 \)

Question 7

In an A.P. sum of three consecutive terms is 27 and their product is 504 find the terms? (Assume that three consecutive terms in A.P. are \( a – d, a, a + d \).)

Answer:

Let the first term be \( a – d \)
the second term be \( a \)
the third term be \( a + d \)

Given: sum of consecutive three term is 27

\( (a – d) + a + (a + d) = 27 \)

\( 3 a = 27 \)

\( a = 9 \)

Also, Given product of three consecutive term is 504

\( (a – d) \times a \times (a + d) = 504 \)

\( (9 – d) \times 9 \times (9 + d) = 504 \) (since, \( a = 9 \))

\( 9^2 – d^2 = 56 \) (since, \( (a – b)(a + b) = a^2 – b^2 \))

\( 81 – d^2 = 56 \)

\( d^2 = 81 – 56 = 25 \)

\( d = \sqrt{25} = \pm 5 \)

Case 1:
Thus, if \( a = 9 \) and \( d = 5 \)
Then the three terms are,
First term \( a – d = 9 – 5 = 4 \)
Second term \( a = 9 \)
Third term \( a + d = 9 + 5 = 14 \)
Thus, the A.P. is 4, 9, 14

Case 2:
Thus, if \( a = 9 \) and \( d = – 5 \)
Then the three terms are,
First term \( a – d = 9 – ( – 5) = 9 + 5 = 14 \)
Second term \( a = 9 \)
Third term \( a + d = 9 + ( – 5) = 9 – 5 = 4 \)
Thus, the A.P. is 14, 9, 4

Question 8

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

Answer:

Let the first term be \( a – d \)
the second term be \( a \)
the third term be \( a + d \)
the fourth term be \( a + 2 d \)

Given: sum of consecutive four term is 12

\( (a – d) + a + (a + d) + (a + 2d) = 12 \)

\( 4 a + 2d = 12 \)

\( 2(2 a + d) = 12 \)

\( 2a + d = 6 .....(1) \)

Also, sum of third and fourth term is 14

\( (a + d) + (a + 2d) = 14 \)

\( 2a + 3d = 14 ......(2) \)

Subtracting eq. (1) from eq. (2) we get,

\( (2a + 3d) – (2a + d) = 14 – 6 \)

\( 2a + 3d – 2a – d = 8 \)

\( 2d = 8 \)

\( d = 4 \)

Substituting value of "d" in eq. (1) we get,

\( 2a + 4 = 6 \)

\( 2a = 6 – 4 = 2 \)

\( a = 1 \)

Thus, \( a = 1 \) and \( d = 4 \)

Hence, first term \( a – d = 1 – 4 = – 3 \)
the second term \( a = 1 \)
the third term \( a + d = 1 + 4 = 5 \)
the fourth term \( a + 2 d = 1 + 2\times4 = 1 + 8 = 9 \)

Thus, the A.P. is – 3, 1, 5, 9

Question 9

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

Answer:

Now, By using nth term of an A.P. formula:

\( t_n = a + (n – 1)d \)

Given: \( t_9 = 0 \)

\( t_9 = a + (9 – 1)d \)

\( 0 = a + 8d \)

\( a = – 8d \)

To Show: \( t_{29} = 2 \times t_{19} \)

Now,

\( t_{29} = a + (29 – 1)d \)

\( t_{29} = a + 28d \)

\( t_{29} = – 8d + 28d = 20 d \) (since, \( a = – 8d \))

\( t_{29} = 20 d \)

\( t_{29} = 2 \times 10 d ....(1) \)

Also,

\( t_{19} = a + (19 – 1)d \)

\( t_{19} = a + 18d \)

\( t_{19} = – 8d + 18d = 10 d \) (since, \( a = – 8d \))

\( t_{19} = 10 d .....(2) \)

From eq. (1) and eq. (2) we get,

\( t_{29} = 2 \times t_{19} \)

Title: Practice Set 3.3 Arithmetic Progression Class 10th Solutions Labels: Class 10 Maths, Arithmetic Progression, MH Board Solutions, Practice Set 3.3 Permanent Link: practice-set-3-3-arithmetic-progression-class-10-solutions Search Description: Detailed step-by-step solutions for Practice Set 3.3 Arithmetic Progression, Class 10 Mathematics Part 1, Maharashtra Board.