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. $81$
• B. $243$
• C. $343$
• D. $324$

Answer 1

Answer: (B)

Given $\frac{ar(r^{10}-1)}{r-1}=18...(1)$
Also $\frac{\frac{1}{ar}(1-\frac{1}{r^{10}})}{1-\frac{1}{r}}=6$
or $\frac{1}{a{r}^{11}}\cdot \frac{(r^{10}-1)r}{r-1}=6$
or $\frac{1}{a^2{r}^{11}}\cdot \frac{ar(r^{10}-1)}{r-1}=6...(2)$
From (1) and (2),
$\frac{1}{a^2{r}^{11}}\times 18=6$
or $a^2{r}^{11}=3$
Now $P=a^{10}{r}^{55}=(a^2{r}^{11}{)}^5$
$=3^5=243$