Question

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

Answer: (C)

Given $\frac{\mathrm{ar}\left(r^{10}-1\right)}{r-1}=18$....(1)
Also $\frac{\frac{1}{\mathrm{ar}}\left(1-\frac{1}{r^{10}}\right)}{1-\frac{1}{r}}=6\Rightarrow \frac{1}{{\mathrm{ar}}^{11}}\cdot \frac{\left(r^{10}-1\right)r}{r-1}=6$
$\frac{1}{a^2{r}^{11}}\cdot \frac{\mathrm{ar}\left(r^{10}-1\right)}{r-1}=6\dots (2)$
$\text{ from (1) and }(2)$
$$\begin{gathered} \frac{1}{a^2{r}^{11}}\cdot 18=6\Rightarrow a^2{r}^{11}=3 \\ <br/>\text{ now }P=a^{10}{r}^{55}={\left(a^2{r}^{11}\right)}^5=3^5=243 \end{gathered}$$