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Question

Let an denote the nth term of a geometric progression with common ratio less than 1. If a1+a2+a3=13 and a21+a22+a23=91, then the value of a10 is

A
310
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B
311
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C
1310
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D
137
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Solution

The correct option is D 137
Let a1=ar,a2=a,a3=ar
a1+a2+a3=13
ar+a+ar=13
a(1r+r)=13a
1r+r=13aa (1)

a21+a22+a23=91
a2(1r2+1+r2)=91
a2[(1r+r)21]=91
a2[(13aa)21]=91 [From (1)]
(13a)2a2=91
a=3

So, 1r+r=103
3r210r+3=0
r=13 [r<1]

a1=9a10=9(13)9=(13)7

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