Question

The sum of infinite terms of the following series $1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+.......$ will be
[MP PET 1981; RPET 1997; Roorkee 1992; DCE 1996, 2000]

• A. $\frac{3}{16}$
• B. $\frac{35}{8}$
• C. $\frac{35}{4}$
• D. $\frac{35}{16}$

Answer 1

The correct option is D $\frac{35}{16}$

Let the sum to infinity of the arithmetico-geometric series be $1+4.\frac{1}{5}+7.\frac{1}{5^2}+10.\frac{1}{5^3}+.......$
$\Rightarrow \frac{1}{5}S=\frac{1}{5}+4.\frac{1}{5^2}+7.\frac{1}{5^3}+.......$
Subtracting $(1-\frac{1}{5})S=1+3.\frac{1}{5}+3.\frac{1}{5^2}+3.\frac{1}{5^3}+.....$
$=1+3(\frac{1}{5}+\frac{1}{5^2}+....)$

$\Rightarrow \frac{4}{5}.S=1+3.\frac{1}{5}\left(\frac{1}{1-\frac{1}{5}}\right)=1+\frac{3}{4}=\frac{7}{4}\Rightarrow S=\frac{35}{16}.$
Aliter : Use direct formula $S_{\infty }=\frac{ab}{1-r}+\frac{dbr}{(1-r{)}^2}$
Here a = 1, b = 1, d = 3, r = \frac{1}{5}, therefore $S_{\infty }=\frac{1}{1-\frac{1}{5}}+\frac{3\times 1\times \frac{1}{5}}{{(1-\frac{1}{5})}^2}=\frac{5}{4}+\frac{\frac{3}{5}}{\frac{16}{25}}=\frac{5}{4}+\frac{15}{16}=\frac{35}{16}.$
Aliter : Use $S=[1+\frac{r}{1-r}\times diff.ofA.P.]\frac{1}{1-r}$