Question

The sum of an infinite geometric series is $2$ and the sum of geometric series made from the cubes of this infinite series is $24$. Then the series is
A. $3+\dfrac{3}{2}-\dfrac{3}{4}+\dfrac{3}{8}-...$
B. $3+\dfrac{3}{2}-\dfrac{3}{4}+\dfrac{3}{8}+...$
C. $3-\dfrac{3}{2}+\dfrac{3}{4}-\dfrac{3}{8}+...$
D. None of these

Answer 1

Hint: In this question, we are to find the sum of an infinite geometric series. Here we have the sum to infinite geometric series and the sum to of geometric series made from the cubes of infinite series. By using them, we get the common ratio and the first term. Thus, we can form a geometric series.

Formula Used: The general geometric series is written as
$a,ar,a{{r}^{2}},a{{r}^{3}},...,a{{r}^{n}}$
The common ratio is
$r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
The sum of the infinite terms in a G.P series is
 ${{S}_{\infty }}=\dfrac{a}{1-r}$
Here ${{S}_{\infty }}$ - The sum of the infinite terms of the series; $a$- The first term and $r$- The common ratio.

Complete step by step solution: Consider a geometric series as
$a, ar, a{{r}^{2}}, a{{r}^{3}},...$
Then the sum of the above infinite series is
${{S}_{\infty }}=\dfrac{a}{1-r}$
When applying cube for the above series, we get a new series as
${{a}^{3}},{{a}^{3}}{{r}^{3}},{{a}^{3}}{{r}^{6}},{{a}^{3}}{{r}^{9}},...$
Then the sum to the infinity of the new series obtained above is
${{S}_{\infty }}=\dfrac{{{a}^{3}}}{1-{{r}^{3}}}$
It is given that,
The sum of the required infinite geometric series is $2$.
I.e., $2=\dfrac{a}{1-r}\text{ }...(1)$
The sum of geometric series made from the cubes of the required infinite series is $24$.
I.e., $24=\dfrac{{{a}^{3}}}{1-{{r}^{3}}}\text{ }...(2)$
On applying cube in (1), we get
$\begin{align}
  & {{2}^{3}}={{\left( \dfrac{a}{1-r} \right)}^{3}} \\
 & \Rightarrow 8=\dfrac{{{a}^{3}}}{{{(1-r)}^{3}}}\text{ }...(3) \\
\end{align}$
On dividing (2) by (3), we get
$\begin{align}
  & \dfrac{24}{8}=\dfrac{\dfrac{{{a}^{3}}}{1-{{r}^{3}}}}{\dfrac{{{a}^{3}}}{{{(1-r)}^{3}}}} \\
 & \Rightarrow 3=\dfrac{{{(1-r)}^{3}}}{1-{{r}^{3}}} \\
\end{align}$
But we know that,
${{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})$
Then,
$\begin{align}
  & 3=\dfrac{{{(1-r)}^{3}}}{1-{{r}^{3}}} \\
 & \Rightarrow 3=\dfrac{{{(1-r)}^{3}}}{(1-r)(1+r+{{r}^{2}})} \\
 & \Rightarrow 3=\dfrac{{{(1-r)}^{2}}}{(1+r+{{r}^{2}})} \\
 & \Rightarrow 3(1+r+{{r}^{2}})={{(1-r)}^{2}} \\
\end{align}$
On simplifying, we get
$\begin{align}
  & \Rightarrow 3(1+r+{{r}^{2}})=1-2r+{{r}^{2}} \\
 & \Rightarrow 3+3r+3{{r}^{2}}=1-2r+{{r}^{2}} \\
 & \Rightarrow 2{{r}^{2}}+5r+2=0 \\
\end{align}$
On factorizing the obtained equation, we get
$\begin{align}
  & 2{{r}^{2}}+5r+2=0 \\
 & \Rightarrow 2{{r}^{2}}+4r+r+2=0 \\
 & \Rightarrow 2r(r+2)+(r+2)=0 \\
 & \Rightarrow (r+2)(2r+1)=0 \\
\end{align}$
Thus, $r=-2$ and $r=\dfrac{-1}{2}$. But we know that $-1On substituting $r=\dfrac{-1}{2}$ in (1), we get
$\begin{align}
  & 2=\dfrac{a}{1-r} \\
 & \Rightarrow a=2(1-r)=2\left( 1+\dfrac{1}{2} \right) \\
 & \therefore a=3 \\
\end{align}$
So, the required series with the first term $a=3$ and the common ratio $r=\dfrac{-1}{2}$ is
$\begin{align}
  & a+ar+a{{r}^{2}}+a{{r}^{3}}+...=3+3\left( \dfrac{-1}{2} \right)+3{{\left( \dfrac{-1}{2} \right)}^{2}}+... \\
 & \Rightarrow 3-\dfrac{3}{2}+\dfrac{3}{4}-\dfrac{3}{8}+... \\
\end{align}$

Option ‘C’ is correct

Note: Here, we need to remember that, for forming a geometric series we need the first term and the common ratio. So, by using the given sums, we find the common ratio and the first term.