Question

1. How many terms of the AP 9, 17,25,... must be taken so that there sum is 636​

Answer 1

${}{ \huge{ \bold{ \underline{ \red{solution - }}}}} \\ \\ \\ {} \huge \green{hope \: it \: helps \: you}$

Answer 2

Answer:

a = 9

d = 17 - 9

= 8

Sn = 636

Sn = n/2 (2a +(n-1)d)

636 = n/2 (2(9) + (n-1) 8)

636 = n/2 x 2 (9 +(n-1)4)

636 = n(9 +4n - 4)

636 = 5n +4n^2

4n^2 +5n - 636 =0

Let's split

4n^2 - 48n +53 n- 636 =0

4n(n - 12) + 53 (n-12) =0

(4n +53) (n-12) =0

4n +53 =0 or n-12 =0

4n =-53 or n =12

n =-53/4 or n =12

So, the n cannot be negative so, our final answer is 12

So, S12 =636