Question

In an A.P S7 = 49,S17= 289 find Sn

Answer 1

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Answer 2

Answer:

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

Step-by-step explanation:

Formula of first n terms in AP = $S_n = \frac{n}{2}(2a+(n-1)d)$

Substitute n = 7

So, $S_7 = \frac{7}{2}(2a+6d)$

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

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

$14= 2a+6d$

Substitute n = 17

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

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

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

$34= 2a+16d$

$34= 2a+10d+6d$

Using A

$34= 14+10d$

$20 = 10d$

$d=2$

Substitute the value of d in A

$14= 2a+12$

$2= 2a$

$a=1$

So,  $S_n = \frac{n}{2}(2(1)+(n-1)2)$

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