Question

The geometric series $a+ar+a{r}^2+a{r}^3+......\infty$ has sum $7$ and the terms involving odd powders of $r$ has sum ${}^{\prime }{3}^{\prime }$, then the value of $(a^2-r^2)$ is-

• A. $\frac{5}{4}$
• B. $\frac{5}{2}$
• C. $\frac{25}{4}$
• D. $5$

Answer 1

The correct option is B $\frac{5}{2}$

Sum of geometric series = $\frac{a(r^n-1)}{r-1}=7$

Also odd powers of r terms :-

$ar,a{r}^3,a{r}^5,a{r}^7$

$\therefore sum=\frac{ar(r^{2n}-1)}{r^2-1}$

$\therefore$ sum of infinite G.P = $\frac{a}{1-r}=7$ _________(1)

$\therefore$ sum of infinite second = $\frac{ar}{1-r^2}=3$ _____(2)

$\therefore$ From (1) & (2)

$\frac{1+r}{r}=\frac{7}{3};r=\frac{3}{4}$

$\therefore a=\frac{7}{4}$

$\therefore a^2-r^2=\frac{5}{2}$