1.

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

Answer»

A=9, d=17-9=8, Sn=636
Sn=n/2 [(2A+(n-1) d]
636= n/2 [ 2*9+(n-1)*8]
636= n/2 [18+8n-8]
636*2= n (10+8n)
1272= 10n+8n^2
8n^2+10n-1272=0
2(4n^2+5n-636)=0
4n^2+5n-636=0
4n^2+53n-48n-636=0
n(4n+53)-12(4n+53)=0
(n-12) (4n+53)=0
n=12, n=-53/4
n=-53/4 is not possible
So, n=12


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