If S is the sum to infinity of a G.P., whose first term is a, then the sum of the first n terms is [UPSEAT 2002]
A
S(1−aS)n
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B
S[1−(1−aS)n]
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C
a[1−(1−aS)n]
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D
None of these
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Solution
The correct option is BS[1−(1−aS)n] Let r be the common ratio of the G.P. Then S=a1−r⇒r=1−aS
Now Sn = Sum of n terms =a(1−rn1−r)=a1−r(1−rn)=S[1−(1−aS)n].