If n GMs be inserted between a and b, then the corresponding common ratio is
A
(ba)1/n+1
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B
(ab)1/n+1
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C
(ba)1/n
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D
(ab)1/n
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Solution
The correct option is A(ba)1/n+1 Let G1,G2,.............Gn are the G.Ms inserted between aandb and the common ration be ′r′ ⇒G1=ar,G2=ar2,.........Gn=arn,b=arn+1 So using last term rn+1=ba⇒r=(ba)1/n+1 Hence, option 'A' is correct.