Question

how many terms of the ap 9, 17, 25....must be taken to give sum of 636?

Answer 1

Sn = n/2 (2a +(n-1)d)
636 = n/2 ( 18 +(n-1)8)
636*2 = n ( 18+8n-8)
636*2 = n ( 10+8n)
1272 = 10n+8n^2
8n^2 + 10n -1272
4n^2 + 5n - 636
4n^2 + 53n - 48n - 636
n(4n+53)-12(4n+53)
so , n = -53/4
and n = 12
  

Answer 2

Sn = n/2 [2a+{n-1}d]
here,
a = 9 , d = 8 and Sn = 636
to find n
from the formula written above
636 = n/2 [2*9 + {n-1}8]
1272 = n[18 + 8n - 8]
1272 = n[ 8n + 10]
1272 = 8n^2 + 10n
4n^2 + 5n - 636 = 0.
4n^2 + 53n - 48n - 636 = 0.
4n^2 - 48n + 53n - 636 = 0
4n [n - 12] + 53 [ n - 12 ] = 0
[ n - 12 ] [ 4n + 53 ] = 0.
therefore,
n = - 53 /4 or n = 12.
but the number of terms can be neither a fraction nor can it be negative.
so n = 12.
thus 12 terms of the given a.p must be taken so that the sum becomes 636.