A geometric progression with common ratio r, consists of an even number of terms. If the sum of all terms is 5 times the sum of the terms occupying the odd places, then 4∑i=1(ir)2 is
A
1456
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B
120
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C
1172
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D
480
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Solution
The correct option is D480 Let the G.P. be a,ar,ar2,…,ar2n−1. Then, S2n=5(a+ar2+⋯+nterms) ⇒a1−r(1−r2n)=5⋅a1−r2(1−(r2)n) ⇒r+1=5 ⇒r=4