Question

If $a_1,a_2,a_3,.......$ are in G.P. with first term 'a' and common ratio 'r', then $\frac{a_1{a}_2}{a_1^2-a_2^2}+\frac{a_2{a}_3}{a_2^2-a_3^2}+\frac{a_3{a}_4}{a_3^2-a_4^2}+.......+\frac{a_{n-1}{a}_n}{a_{n-1}^2-a_n^2}$ is equal to

• A. $\frac{nr}{1-r^2}$
• B. $\frac{(n-1)r}{1-r^2}$
• C. $\frac{nr}{1-r}$
• D. $\frac{(n-1)r}{1-r}$

Answer 1

Answer: (B)

$\frac{a_1{a}_2}{a_1^2-a_2^2}+\frac{a_2{a}_3}{a_2^2-a_3^2}+\frac{a_3{a}_4}{a_3^2-a_4^2}+.....+\frac{a_{n-1}{a}_n}{a_{n-1}^2-a_n^2}$
$=\frac{a_1.a_{1}r}{a_1^2\left(1-r^2\right)}+\frac{a_2.a_{2}r}{a_2^2\left(1-r^2\right)}+\frac{a_3.a_{3}r}{a_3^2\left(1-r^2\right)}+.....+\frac{a_{n-1}.a_{n-1}r}{a_{n-1}^2\left(1-r^2\right)}$
$=\frac{r}{1-r^2}+\frac{r}{1-r^2}+.....+\frac{r}{1-r^2}$ (no. of terms $n-1$)
$=\frac{\left(n-1\right)r}{1-r^2}$