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Question

A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of the terms occupying the odd places. Find the common ratio of the G.P.

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Solution

Let the G.P. be 2n,2,2n+4,.....

Then, Sn=a(rn1)r1,a=2n,r=2

Sn=2n(2n1)21=2nn+12n

Then the G.P. of odd term

a1+a3+a5+.....a2n1

According to the question

Sum of all terms = 5 (sum of terms occupying the odd places)

a1+a2+a3+.....+a2n=5(a1+a3+a5+....a2n1)

a+ar+ar2+....+ar2n1=5(a+ar2+ar4+....+ar2n2)

a(1r2n)1r=5(a(1(r)2)n1r2)

Take (a1r) common from both sides

1r2n=5(1r2n)1+r

1+rr2nr2n+1=55r2n

r2n+14r2nr+4=0

r2n(r4)1(r4)=0

(r2n1)(r4)=0

either r2n=1, r=4

r=4


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