Question

If $A=1+r^z+r^{2z}+r^{3z}+....\infty$, then the value of $r$ will be

• A. $A{(1-A)}^z$
• B. ${\left(\frac{A-1}{A}\right)}^{\frac{1}{Z}}$
• C. ${\left(\frac{1}{A-1}\right)}^{\frac{1}{Z}}$
• D. $A{(1-A)}^{\frac{1}{Z}}$

Answer 1

The correct option is B

${\left(\frac{A-1}{A}\right)}^{\frac{1}{Z}}$

Find the value of $r$ :

Given, $A=1+r^z+r^{2z}+r^{3z}+....\infty$

This is a G.P

Here, $a=1$

common ratio, $r=r^z$

$\Rightarrow$$A=\frac{1}{1-r^z}$ $\left[\mathbf{\because }{\mathit{S}}_{\mathbf{\infty }}\mathbf{=}\mathbf{}\frac{\mathbf{a}}{\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathbf{r}}\right]$

$\Rightarrow$$A-A{r}^z=1$

$\Rightarrow$$A{r}^z=A-1$

$\Rightarrow$$r^z=\frac{A-1}{A}$

$\mathbf{\therefore }r={\left(\frac{\mathrm{A}-1}{\mathrm{A}}\right)}^{\frac{1}{\mathrm{Z}}}$

Hence, Option ‘B’ is Correct.