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Question

Find the sum of n terms of the series loga+loga2b+loga3b2+loga4b3+.... to n terms .

A
n4[nlogba+logab]
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B
n4[nlogab+logab]
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C
n2[nlogab+logab]
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D
n8[nlogba+logab]
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Solution

The correct option is C n2[nlogab+logab]
TnTn1=loganbn1logan1bn2
loganbn1bn2an1=logab= constant as it is independent of n.
Hence the given series is an A.P. whose A=loga and D=logab

S=n2[2loga+[n1]logab] =n2[nlogab+2loga(logalogb)]
=n2[nlogab+logab]
Hence, option C.

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