#### 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, S_{7} = 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, S_{17} = 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

⇒ d = 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 n^{2}.

Is there an error in this question or solution?

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.