How many terms of the AP: 9, 17, 25, . . . must be taken to give a sum of 636?

Explanation

d=8

a=9

`S_n=n/2(2a+(n-1)d)`

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

`4n^2+5n-636`

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

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

n=12,-53/4

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