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If `sin theta,sqrt(2) (sintheta + 1),6 sin theta +6` are in GP, than the fifth term is
A. 81
B. `81sqrt(2)`
C. 162
D. `162sqrt(2)`

1 Answer

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Best answer
`[sqrt(2)(sintheta + 1)]^(2)= sin theta (6 sin theta +6)`
`implies [(sintheta + 1) 2(sin theta+1)-6 sin theta]=0`
We get, `sin theta=-1, (1)/(2)`
` therefore sin theta=(1)/(2)" [" sin theta =-1 " is not possible ]"`
then first term` =a=sin theta=(1)/(2)` and common ratio
`=r= sqrt(2)((1)/(2)+1)/((1)/(2))=3sqrt(2)`
`therefore t_(5)=ar^(4)=(1)/(2)(3sqrt(2))^(4)=162`
Hence, (c) is the correct answer.

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