Question

Sum to infinite terms of the GP $\frac{2}{5},\frac{3}{5^2},\frac{2}{5^3},\frac{3}{5^4}..$is ___.

• A. $\frac{7}{24}$
• B. $\frac{5}{24}$
• C. $\frac{13}{24}$
• D. $\frac{11}{24}$

Answer 1

The correct option is C

$\frac{13}{24}$

$\frac{2}{5}+\frac{3}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+\frac{2}{5^5}+\frac{3}{5^6}\cdots \cdots$

$=\{\frac{2}{5}+\frac{2}{5^3}+\frac{2}{5^5}\cdots \}+\{\frac{3}{5^2}+\frac{3}{5^4}+\frac{3}{5^6}\cdots \}$

$=\frac{2}{5}\{1+\frac{1}{5^2}+\frac{1}{5^4}\cdots \}+\frac{3}{5^2}\{1+\frac{1}{5^2}+\frac{1}{5^4}\cdots \}$

$=(\frac{2}{5}+\frac{3}{5^2})\{1+\frac{1}{5^2}+\frac{1}{5^4}\cdots \}$

$=(\frac{2}{5}+\frac{3}{25})\left[\frac{1}{1-\frac{1}{25}}\right](a=1,r=\frac{1}{25})$

$=\frac{13}{25}\times \frac{25}{24}$

$=\frac{13}{24}$