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Question

If G1,G2,G3....Gn are n numbers inserted between the numbers a and b of a G.P. then G1×G2×G3×....Gn=

A
ab
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B
(ab)n
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C
(ab)n+1
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D
(a+b2)n
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Solution

The correct option is B (ab)n
let a,G1,G2,G3,...Gn,b be in G.P
G1=arG2=ar2....Gn=arn
G1.G2........Gn=(an).(rn(n+1)2)[1]
and a×b=(a2).(r(n+1))
finding r from second equation in terms of ab and substituting it in first equation gives
G1.G2........Gn=(ab)n

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