Question

Consider the ten numbers $ar , ar ^2, ar ^3, \ldots \ldots \ldots \ldots ar ^{10}$. If their sum is 18 and the sum of their reciprocals is 6 then the product of these ten numbers, is

• A. 324
• B. 343
• C. 243
• D. 729

Answer 1

Answer: (C)

Given $\frac{\operatorname{ar}\left(r^{10}-1\right)}{r-1}=18$....(1)
Also $\frac{\frac{1}{\operatorname{ar}}\left(1-\frac{1}{r^{10}}\right)}{1-\frac{1}{r}}=6 \Rightarrow \frac{1}{\operatorname{ar}^{11}} \cdot \frac{\left( r ^{10}-1\right) r }{ r -1}=6$
$ \frac{1}{a^2 r^{11}} \cdot \frac{\operatorname{ar}\left(r^{10}-1\right)}{r-1}=6 \ldots(2) $
$\text { from (1) and }(2) $
$ \frac{1}{a^2 r^{11}} \cdot 18=6 \Rightarrow a^2 r^{11}=3 \\
\text { now } P=a^{10} r^{55}=\left(a^2 r^{11}\right)^5=3^5=243$