Question

Consider $10$ numbers $ar,a{r}^2,a{r}^3,....,a{r}^{10}$. Their sum is $18$ and the sum of their reciprocal is $6$. If the product of these $10$ numbers is $a^b$ (where $a$ and $b$ are prime) then find the value of $a+b$.

Answer 1

Given

$ar+a{r}^2+a{r}^3+......a{r}^{10}=18$ …………..$\left(1\right)$

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

From $\left(2\right)$

$\frac{(r^9+r^8+.....1)}{a{r}^{10}}=6$

$\frac{ar(r^9+r^8+......1)}{ar\left(a{r}^{10}\right)}=6$

$\frac{a{r}^{10}+a{r}^9+....ar}{a^2{r}^{11}}=6$

$\Rightarrow \frac{\text{/}{18}^3}{a^2{r}^{11}}=\text{/}6$

$\Rightarrow a^2{r}^{11}=3$

Product of $10$ numbers $=\left(ar\right)\left(a{r}^2\right)\left(a{r}^3\right)\dots .\left(a{r}^{10}\right)$

$=a^{10}{r}^{(1+\mathrm{2....10})}=a^{10}{r}^{55}$

$=(a^2{r}^{11}{)}^5\Rightarrow 3^5$

$\therefore$ The value of $a+b=3+5=8$.