Question

The sum of first $20$ terms of the sequence $0.7,0.77,0.777,....$ is:
A)$\dfrac{7}{81}\left( 179-{{10}^{-20}} \right)$
B)$\dfrac{7}{9}\left( 99-{{10}^{-20}} \right)$
C)$\dfrac{7}{81}\left( 179+{{10}^{-20}} \right)$
D)$\dfrac{7}{9}\left( 99+{{10}^{-20}} \right)$

Answer 1

Hint: he question can be solved using the concept of Geometric Progression (G.P) where the sum of Geometric Progression (G.P) is given by $\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r},r<1$ where $a$ is the first term and $r$ is the common ratio, this formula is applicable only if the common ratio of the series is less than one. In a Geometric Progression (G.P) the terms are arranged such that they have a common ratio and which is given by $a+ar+a{{r}^{2}}+a{{r}^{3}}+.....a{{r}^{n}}$.

Complete step by step solution:
The sum of the terms of sequence is represented by:
$S=0.7+0.77+0.777+.........\left( 20terms \right)$……(1)
Eliminating the common term $7$ and rewriting the equation (1) we get:
$\Rightarrow S=7\left( 0.1+0.11+0.111+.........\left( 20terms \right) \right)$ ……(2)
Multiplying and dividing by $9$ in equation (2) we get:
$\Rightarrow S=\dfrac{7}{9}\left( 0.9+0.99+0.999+.........\left( 20terms \right) \right)$ ……(3)
Writing $0.9=\left( 1-0.1 \right)$ so that we can get a simplified version in equation (3) we get:
\[\Rightarrow S=\dfrac{7}{9}\left( \left( 1-0.1 \right)+\left( 1-0.01 \right)+\left( 1-0.001 \right)+.........20terms \right)\] ……(4)
Since there are $20$ terms of $1$ therefore equation (4) can be rewritten and we get:
$\Rightarrow S=\dfrac{7}{9}\left[ 20-\left( \dfrac{1}{10}+\dfrac{1}{{{10}^{2}}}+\dfrac{1}{{{10}^{3}}}+........\dfrac{1}{{{10}^{20}}} \right) \right]$ ……(5)
The terms $\dfrac{1}{10}+\dfrac{1}{{{10}^{2}}}+\dfrac{1}{{{10}^{3}}}+........\dfrac{1}{{{10}^{20}}}$ forms a Geometric Progression (G.P) with first term $\dfrac{1}{10}$ and common ratio $\dfrac{1}{10}$ .
Sum of Geometric Progression (G.P) is given by $\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r},r<1$
Here the values of a and r are as listed below:
$\begin{align}
  & a=\dfrac{1}{10},r=\dfrac{1}{10} \\
 & S=\dfrac{7}{9}\left[ 20-\dfrac{1}{10}\left( \dfrac{1-{{\left( \dfrac{1}{10} \right)}^{n}}}{1-\dfrac{1}{10}} \right) \right] \\
\end{align}$ ……(6)
$\begin{align}
  & \Rightarrow S=\dfrac{7}{9}\left[ 20-\dfrac{1}{10}\left( \dfrac{1-{{10}^{-20}}}{\dfrac{9}{10}} \right) \right] \\
 & \Rightarrow S=\dfrac{7}{9}\left[ 20-\dfrac{1}{9}\left( 1-{{10}^{-20}} \right) \right] \\
 & \Rightarrow S=\dfrac{7}{9}\left[ \dfrac{180-\left( 1-{{10}^{-20}} \right)}{9} \right] \\
\end{align}$

On simplifying the above equation we get:
$\therefore S=\dfrac{7}{81}\left[ 179+{{10}^{-20}} \right]$

So, the correct answer is “Option C”.

Note: Since here the number of terms is specified therefore the sum of infinite series of Geometric Progression (G.P) should not be used i.e. $\dfrac{a}{1-r}$ and the common ratio is less than $1$ therefore the correct formula of Geometric Progression (G.P) should be used i.e. $\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r},r<1$.
The formula $\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1},r<1$ should not be used.