Question

The sum of the 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

$\frac{35}{16}$

Explanation for the correct option:

Let the sum be $S$

$S=1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}......\left(1\right)$
$\Rightarrow \frac{S}{5}=\frac{1}{5}+\frac{4}{5^2}+\frac{7}{5^3}+\frac{10}{5^4}......\left(2\right)$

Subtracting $\left(2\right)$ from $\left(1\right)$,

$$\begin{gathered} \Rightarrow \frac{4S}{5}=1+\frac{3}{5}+\frac{3}{5^2}+\frac{3}{5^3}+\dots \\ \Rightarrow \frac{4S}{5}=1+3\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\dots \right) \end{gathered}$$

We can see that the terms in the bracket in $RHS$ excluding the $1$ is in geometric progression.
The sum of a geometric series (geometric series is a sum whose terms are in geometric progression) with infinite terms and common ratio less than $1$ is given as,

$S_n=\frac{a}{1-r}$
where $a$ is the first term and $r$ is the common ratio.

Here, $a=\frac{1}{5}$ and $r=\frac{1}{5}$

Thus, How? Explain the formula used

$\begin{array}{cc}\Rightarrow & \frac{4S}{5}=1+3\left(\frac{{\displaystyle \frac{1}{5}}}{1-{\displaystyle \frac{1}{5}}}\right)\\ \Rightarrow & \frac{4S}{5}=1+\frac{3}{4}\\ \Rightarrow & S=\frac{7}{4}\times \frac{5}{4}\\ \Rightarrow & S=\frac{35}{16}\end{array}$

Therefore, the sum of the infinite series is $\frac{35}{16}$.

Hence, the correct option is (D).