Question

the sum of how many terms of the AP 9,17,25,... is 636?

Answer 1

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

here a=9. , d=8. , Sn=636 , n=??

636 = n/2[18 + 8n -8 ]

1272 = 18n + 8n² -8n

1272 = 8n² + 10n

636 = 4n² + 5n

4n² + 5n - 636 =0
4n² + 53n -48n -636 =0
n(4n+53)-48(4n+53)=0
(n-48) (4n+53)=0
then

n=48 or n=(-53/4)

n cannot be negative or fraction so n=48

hence sum of 48 terms of given A.P is 636

HOPE IT HELPS

Answer 2

Let a=first term

d=difference

Given a=9

d=17-9

=8

Sn=636

n/2*[2a+(n-1)d]

636=n/2*[18+(n-1)8]

1272=n[18+8n-8]

=1272=n[10+8n]

=1272=10n+8n²

=4n²+5n-636=0

=4n²-48n+53n-636=0

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

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

n=12 or n=-53/4

but n cannot be negative.

n=12.

12 terms need to be added.

Hope it helps.