Question

Class 10



How many terms of the AP : 9,17,25....

must be taken to give a sum of 636?


Answer - 12​

Answer 1

Hey Mate...

$636 = \frac{n}{2} (2a + (n - 1)d)) \\ 636 = \frac{n}{2} (18 + (n - 1)8)$

$636 = n(9 + 4n - 4) \\ 636 = 5n + 4 {n}^{2} \\ 4 {n}^{2} + 5n - 636 = 0$

So...

We get..

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

Hence no.of terms is 12.

Hope it helped ❤️

Answer 2

a=9

d=8

Sn=636

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

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

1272=n{18+8n-8}

1272=n(10+8n)

1272=10n +8n^2. Divide by 2 we get,

636=5n+4n^2

4n^2+5n-636