Question

Consider the ten numbers
$ar,a{r}^2,a{r}^3,\dots \dots \dots .a^{10}$.
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 $\frac{\mathrm{ar}\left(r^{10}-1\right)}{r-1}=18\dots$..(1)
Also $\frac{\frac{1}{ar}\left(1-\frac{1}{r^{10}}\right)}{1-\frac{1}{r}}=6$
$\Rightarrow \frac{1}{a{r}^{11}}\cdot \frac{\left(r^{10}-1\right)r}{r-1}=6$...(2)
from $(1)$ and $(2)$
$\frac{1}{a^2{r}^{11}}\cdot \frac{ar\left(r^{10}-1\right)}{r-1}=6$
$\frac{1}{a^2{r}^{11}}\cdot 18=6$
now $\Rightarrow a^2{r}^{11}=3$
$P=a^{10}{r}^{55}={\left(a^2{r}^{11}\right)}^5=3^5=243$