The sum of first 7 terms of an AP is 49 and that of 17 terms is 289. Find the sum of first n terms .

Given:  The sum of first 7 terms of an AP is 49 and that of 17 terms is 289

To do: Find the sum of first n terms.

Solution:
$S_7 ​ =49$ and $S_17 ​ =289$

By using $S_n ​ = \frac{n}{2} ​ [2a+(n−1)d]$ we have,

$S_7 ​ = \frac{7}{2} ​ [2a+(7−1)d]=49$

⇒$49= \frac{7}{2} ​ [2a+(7−1)d]$

⇒$49= \frac{7}{2} ​ (2a+6d)$

⇒$7=a+3d$

⇒$a+3d=7$...................(i)

$S_17 ​ = \frac{17}{2} ​ [2a+(17−1)d]=289$

⇒$289= \frac{17}{2} ​ [2a+(17−1)d]$

⇒$289= \frac{17}{2} ​ (2a+16d)$

⇒$17=a+8d$

⇒$a+8d=17$......................(ii)

Substituting (i) from (ii), we get

$5d=10$ or $d=2$

From equation (i),

$a+3(2)=7$

$a+6=7$ or $a=1$

$S_n ​ = \frac{n}{2} ​ [2(1)+(n−1)2]$

=$ \frac{n}{2} ​ [2+(n−1)2]$

= $\frac{n}{2} (2+2n−2)=n^2$

Therefore, the sum of first n terms is $n^2$.