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Consider the ten numbers `a r ,a r^2, a r^3, ,a r^(10)dot`If their sum is 18 and the sum of their reciprocals is 6, then theproduct of these ten numbers is`81`b. `243`c. `343`d. 324A. 81B. 243C. 343D. 324 |
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Answer» Correct Answer - B Given `(ar(r^(10)-1))/(r-1)=18` (1) Also `(1/(ar)(1-1/r^(10)))/(1-1/r)=6` or `1/(ar^(11))cdot((r^(10)-1)r)/(r-1)=6` or `1/(a^(2)r^(11))cdot(ar(r^(10)-1))/(r-1)=6` From (1) and (2), `1/(a^(2)r^(11))xx18=6` or `a^(2)r^(11)=3` Now P`=a^(10)^(55)=(a^(2)r^(11))^(5)=3^(5)=243` |
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