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

Given:

The sum of 7 terms of an A.P. is 49 and that of 17 terms is 289.

To do:

We have to find the sum of $n$ terms.

Solution:

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

We know that,

Sum of $n$ terms$ S_{n} =\frac{n}{2}(2a+(n-1)d)$

$S_{7}=\frac{7}{2}[2(a)+(7-1)d]$

$49=\frac{7}{2}(2a+6d)$

$49=7(a+3d)$

$a+3d=7$

$a=7-3d$......(i)

$S_{17}=\frac{17}{2}[2(a)+(17-1)d]$

$289=\frac{17}{2}(2a+16d)$

$289=17(a+8d)$

$a+8d=17$

$7-3d+8d=17$       (From (i))

$5d=17-7$

$d=\frac{10}{5}$

$d=2$

This implies,

$a=7-3(2)$

$a=7-6$

$a=1$

The sum of $n$ terms $S_n=\frac{n}{2}(2a+(n-1)d)$

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

$=n(1+n-1)$

$=n^2$

Hence, the sum of $n$ terms is $n^2$.