Consider the ten numbers ar, ar², ar³, ..., ar¹⁰. If their sum is 18 and the sum of their reciprocals is 6, then the product of these ten numbers is

Given $\frac{\mathrm{ar}\left({\mathrm{r}}^{10}-1\right)}{\mathrm{r}-1}=18$               (1)

Also $\frac{\frac{1}{\mathrm{ar}}\left(1-\frac{1}{{\mathrm{r}}^{10}}\right)}{1-\frac{1}{\mathrm{r}}}=6$

$\begin{array}{l}\mathrm{or} \frac{1}{{\mathrm{ar}}^{11}}\cdot \frac{\left({\mathrm{r}}^{10}-1\right)\mathrm{r}}{\mathrm{r}-1}=6\\ \mathrm{or} \frac{1}{{\mathrm{a}}^2{\mathrm{r}}^{11}}\cdot \frac{\mathrm{ar}\left({\mathrm{r}}^{10}-1\right)}{\mathrm{r}-1}=6 \left(2\right)\end{array}$

From (1) and (2),

$\frac{1}{{\mathrm{a}}^2{\mathrm{r}}^{11}}\times 18=6$

or ${\mathrm{a}}^2{\mathrm{r}}^{11}=3$

Now $\mathrm{P}={\mathrm{a}}^{10}{\mathrm{r}}^{55}={\left({\mathrm{a}}^2{\mathrm{r}}^{11}\right)}^5=3^5=243$