Question

If 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.

Answer 1

Given that, $S_7 = 49$ and $S_{17} = 289$
$S_n = \frac n2 [2a + (n-1)d]$

$S_7 = \frac 72 [2a + (7-1)d]$

$49 = \frac {7}{2} (2a + 6d)$
$7 = (a + 3d)$
$a + 3d = 7 ……..(i)$
Similarly,
$S_{17 }= \frac {17}{2} [2a + (17-1)d]$

$289 = \frac {17}{2} (2a + 16d)$
$17 = (a + 8d)$
$a + 8d = 17 …….(ii)$
Subtracting equation $(i)$ from equation $(ii)$,
$5d = 10$
$d = 2$
From equation (i),
$a + 3(2) = 7$
$a + 6 = 7$
$a = 1$
$Sn = \frac n2 [2a + (n-1)d]$

$Sn= \frac n2 [2(1) + (n-1)(2)]$

$S_n= \frac n2 (2 + 2n - 2)$

$S_n= \frac n2 (2n)$

$S_n= n^2$