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