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 $n^2$. 

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Updated on: 10-Oct-2022

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