Question

The sum of infinite terms of the following series

1 + $\frac{4}{5}$ + $\frac{7}{5^2}$ + $\frac{10}{5^3}$ + ...........∞ will be

• 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 arithmetic-geometric series be

S = 1 + 4 . $\frac{1}{5}$ + 7. $\frac{1}{5^2}$ + 10. $\frac{1}{5^3}$ +........

⇒ $\frac{1}{5}$ 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($\frac{1}{5}$ + $\frac{1}{5^2}$ + ..............)

⇒ $\frac{4}{5}$ S = 1 + 3.$\frac{1}{5}$($\frac{1}{1-\frac{1}{5}}$) = 1 + $\frac{3}{4}$ = $\frac{7}{4}$ ⇒ 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 + $\frac{r}{1-r}$ × diff. of A.P. ]$\frac{1}{1-r}$