Question

how many terms of the ap 9 17 25 must be taken to give a sum of 636​

Answer 1

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

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

636×2=n (18+8n-8)

1272=10n+8n^2

8n^2+10n-1272=0÷2

4n^2+5n-636=0

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

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

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

n-12=0 4n+53=0

n=12 n=-53/4

n=-53/4 is discarded

12 terms of the ap 9 17 25 must be taken to give a sum of 636

Answer 2

Suppose first term be a  

Common Difference = 17 - 9 = 8  

Sum is 636

Hence

$\bf\huge => S_{n} = 636$

$\bf\huge => \frac{n}{2}[2a + (n - 1)d] = 636$

$\bf\huge => \frac{n}{2}[2\times 9 + (n - 1)8] = 636$

$\bf\huge => \frac{n}{2} (8n - 10) = 636$

⇒ n(4n + 5) = 636

⇒ 4n^2 + 5n + 636 = 0  

$\bf\huge => n = \frac{-5 + \sqrt{25 - 4\times 4\times -636}}{2\times 4}$

$\bf\huge = \frac{-5 + \sqrt{25 + 10176}}{8}$

$\bf\huge = \frac{- 5 + \sqrt{10201}}{8}$

$\bf\huge = \frac{-5 + 101}{8}$

$\bf\huge = \frac{96}{8},\frac{-106}{8}$

$\bf\huge = 12 , \frac{-53}{4}$

In this situation “n” cannot be negative  

n = 12