Question

Consider the ten numbers
$ar,a{r}^2,a{r}^3,....a{r}^{10}$
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

The correct option is C 243

Given numbers are $ar,a{r}^2,a{r}^3,...,a{r}^{10}$

Given sum of the numbers $=18$

$\Rightarrow ar+a{r}^2+a{r}^3+...+a{r}^{10}=18$

$\Rightarrow ar(1+r+r^2+...+r^9)=18$ .........$\left(1\right)$

Given sum of their reciprocals is $6$,

$\Rightarrow \frac{1}{ar}+\frac{1}{a{r}^2}+...+\frac{1}{a{r}^{10}}=6$

$\Rightarrow \frac{1}{a{r}^{10}}(1+r+r^2+...+r^9)=6$ ........$\left(2\right)$

Dividing equation$\left(1\right)$ by $\left(2\right)$, we get $a^2{r}^{11}=3$

Product of these ten numbers,

$a\times ar\times a{r}^2\times a{r}^3\times ...\times a{r}^{10}$

$=a^{10}{r}^{55}$

$={\left(a^2{r}^{11}\right)}^5$

$=3^5=243$