Question

If $a_1,a_2,a_3,.......$ are in G.P. with first term 'a' and common ratio 'r', then $\frac{a_1a_2}{a_1^2-a_2^2}+ \frac{a_2a_3}{a_2^2-a_3^2}+\frac{a_3a_4}{a_3^2-a_4^2}+.......+\frac{a_{n-1}a_n}{a_{n-1}^2-a_n^2}$ 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}}$