Question

If $A=1+r^a+r^{2a}+r^{3a}......\infty$ and $B=1+r^b+r^{2b}......\infty$ then$\frac{a}{b}=$

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

Answer 1

Answer: (B)

$\frac{\log \left(\frac{A-1}{A}\right)}{\log \left(\frac{B-1}{B}\right)}$

If $\left|Commonratio\right|<1$, then sum of infinite terms of G.P. is

$S=\frac{FirstTerm}{1-CommonRatio}$

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

Common ratio$=\frac{r^a}{1}=r^a$

$A=\frac{1}{1-r^a}$

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

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

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

Take $\log$ on both

$\log \left(r^a\right)=\log \left(\frac{A-1}{A}\right)$

As $\log \left(m^n\right)=n\log \left(m\right)$

$\Rightarrow a\log \left(r\right)=\log \left(\frac{A-1}{A}\right)$ ....$\left(i\right)$

$B=1+r^b+r^{2b}+r^{3b}+...$

Common ratio$=\frac{r^b}{1}=r^b$

$B=\frac{1}{1-r^b}$

$\Rightarrow 1-r^b=\frac{1}{B}$

$\Rightarrow r^b=1-\frac{1}{B}$

$\Rightarrow r^b=\frac{B-1}{B}$

Take $\log$ on both

$\log \left(r^b\right)=\log \left(\frac{B-1}{B}\right)$

As $\log \left(m^n\right)=n\log \left(m\right)$

$\Rightarrow b\log \left(r\right)=\log \left(\frac{B-1}{B}\right)$ ....$\left(ii\right)$

$\frac{\left(i\right)}{\left(ii\right)}=\frac{a\log r}{b\log r}=\frac{\log \left(\frac{A-1}{A}\right)}{\log \left(\frac{B-1}{B}\right)}$

$\therefore \frac{a}{b}=\frac{\log \left(\frac{A-1}{A}\right)}{\log \left(\frac{B-1}{B}\right)}$