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 the product of these ten numbers is `81` b. `243` c. `343` d. 324

Options

A. 81
B. 243
C. 343
D. 324

Answer

B

Explanation

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`