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