1.

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

Answer» 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
n can not be negative
so, n=12


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