If first term of a decreasing infinite G.P. is 1 and sum is S, then sum of squares of its terms is-

Question

If first term of a decreasing infinite G.P. is 1 and sum is S, then sum of squares of its terms is- (A) S2 (B) 1 / S 2 (C) S2 /(2 S-1) (D) S2 /(2 S+1)

Tardigrade Admin · Accepted Answer

First term of infinite G.P. =1 and sum =5 Let common ratio is r So S=(1/1-r) Let sum of squares of its terms is S prime. ∴ S prime=1+r2+r4 ldots ldots ∞ ∴ S prime=(1/1-r2)=(1/(1+r)(1-r))=(S/(1+(S-1)S)) [∵ S=(1/1-r)] ∴ S prime=(S/((2 S-1)S))=(S2/2 S-1)

Q. If first term of a decreasing infinite G.P. is 1 and sum is $S$ , then sum of squares of its terms is-

$S^{2}$

$1/ S^{2}$

$S^{2} / (2 S - 1)$

$S^{2} / (2 S + 1)$

Q. If first term of a decreasing infinite G.P. is 1 and sum is S, then sum of squares of its terms is-

S2

1/S2

S2/(2S−1)

S2/(2S+1)

First term of infinite G.P. =1 and sum =5 Let common ratio is r So S=1−r1​ Let sum of squares of its terms is S′. ∴S′=1+r2+r4……∞ ∴S′=1−r21​=(1+r)(1−r)1​=(1+SS−1​)S​[∵S=1−r1​] ∴S′=(S2S−1​)S​=2S−1S2​

Q. If first term of a decreasing infinite G.P. is 1 and sum is $S$ , then sum of squares of its terms is-

$S^{2}$

$1/ S^{2}$

$S^{2} / (2 S - 1)$

$S^{2} / (2 S + 1)$