Question

Consider the ten numbers $ar,a{r}^2,ar3,..........a{r}^{10}$.
If their sum is $18$ and the sum of their reciprocal is $6$ then the product of these ten numbers, is

• A. $3^5$
  
• B. $3^8$
• C. $3^{1}0$
• D. $3^{1}5$

Answer 1

Answer: (A)

$3^5$

$ar,a{r}^2,a{r}^3,............a{r}^{10}$ is a GP with first term as $ar$ and common ration as $r$

$=>su{m}_1=ar+a{r}^2+a{r}^3=.......ar+10=18$

$=>su{m}_2=\frac{1}{ar}+\frac{1}{a{r}^2}+...........+\frac{1}{ar10}=6$

multiply $su{m}_2$ by $a^{2}r11$

i.e.,

$=>ar+a{r}^2+a{r}^3+..........+a{r}^{10}=6\times a^2{r}^{11}$

$=>su{m}_1=6a^2{r}^{10}$

$=>18=6a^2{r}^{10}$

$=>a^2{r}^{11}=3$

Now product of the ratio terms of the GP

$=>a^{10}{r}^{1+2+3+......+10}$

$=>a^{10}{r}^{10\times 11}$

$=>a^{10}{r}^{55}={\left(a^2{r}^{11}\right)}^5$

$=>3^5$