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If  the sum of first 7 terms of an A.P. is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P. - CBSE Class 10 - Mathematics

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Question

If the sum of first 7 terms of an A.P. is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P.

Solution

Let the first term and the common difference of the given AP be a and d, respectively.

Sum of the first 7 terms, S7 = 49

We know

`S=n/2[2a+(n-1)d]`

`=>7/2(2a+6d)= 49`

`=>7/2xx2(a+3d)=49`

⇒ a+3d=7             .....(1)

Sum of the first 17 terms, S17 = 289

`=>17/2(2a+16d)=289`

`=>17/2xx2(a+8d)=289`

`=>a+8d=289/17=17`

⇒ a+8d=17           .....(2)

Subtracting (2) from (1), we get

5d=10

⇒ 2

Substituting the value of d in (1), we get

a = 1

Now,

Sum of the first n terms is given by

`S_n=n/2[2a+(n-1)d]`

`=n/2[2xx1+2(n-1)]`

= `n(1+n-1)=n^2`

Therefore, the sum of the first n terms of the AP is n2.

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 NCERT Solution for Mathematics Textbook for Class 10 (2019 to Current)
Chapter 5: Arithmetic Progressions
Q: 9 | Page no. 113
Solution If  the sum of first 7 terms of an A.P. is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P. Concept: Sum of First n Terms of an AP.
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