Question

The sum of an infinite geometric series is 3. A series, which is formed by squares of its terms, have the sum also 3. then the first series will be
A. \[\dfrac{3}{2},\dfrac{3}{4},\dfrac{3}{8},\dfrac{3}{16},...\]
B. \[\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},\dfrac{1}{16},...\]
C. \[\dfrac{1}{3},\dfrac{1}{9},\dfrac{1}{27},\dfrac{1}{81},...\]
D. \[1,-\dfrac{1}{3},\dfrac{1}{{{3}^{2}}},-\dfrac{1}{{{3}^{3}}},...\]

Answer 1

Hint: In this question, we are to find the common ratio of the first series. Since the sums are equal for the given series, the required common ratio is obtained by equating both the sums. Then, we can easily frame the series.

Formula Used:The sum of the infinite terms in G.P series is calculated by
 ${{S}_{\infty }}=\dfrac{a}{1-r}$ where $r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{\infty }}$ is the sum of the infinite terms of the series; $a$ is the first term in the series, and $r$ is the common ratio.



Complete step by step solution:Consider a series of terms which are in G.P as
$a,ar,a{{r}^{2}},.....\infty $
Thus, the sum of the infinite terms of the series is calculated by the formula,
${{S}_{\infty }}=\dfrac{a}{1-r}$
On substituting,
\[{{S}_{\infty }}=\dfrac{a}{1-r}\]
But it is given that ${{S}_{\infty }}=3$.
So,
$\begin{align}
  & 3=\dfrac{a}{1-r} \\
 & \Rightarrow a=3(1-r)\text{ }...(1) \\
\end{align}$
On squaring each term in the first series, we get
${{a}^{2}},{{(ar)}^{2}},{{(a{{r}^{2}})}^{2}},.....\infty $
Since the series is a geometric series, the common ratio is
$r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
\[\begin{align}
  & \Rightarrow r=\dfrac{{{a}_{2}}}{{{a}_{1}}} \\
 & \text{ =}\dfrac{{{a}^{2}}{{r}^{2}}}{{{a}^{2}}} \\
 & \text{ }={{r}^{2}} \\
\end{align}\]
Thus, the sum of the infinite terms of the series is calculated by the formula,
${{S}_{\infty }}=\dfrac{a}{1-r}$
On substituting $a={{a}^{2}};r={{r}^{2}}$, we get
\[{{S}_{\infty }}=\dfrac{{{a}^{2}}}{1-{{r}^{2}}}\]
But it is given that ${{S}_{\infty }}=3$.
So,
$\begin{align}
  & 3=\dfrac{{{a}^{2}}}{1-{{r}^{2}}} \\
 & \Rightarrow {{a}^{2}}=3(1-{{r}^{2}}) \\
\end{align}$
On substituting (1), we get
$\begin{align}
  & {{\left( 3(1-r) \right)}^{2}}=3(1-{{r}^{2}}) \\
 & \Rightarrow 9{{(1-r)}^{2}}=3(1-r)(1+r) \\
 & \Rightarrow 3(1-r)(1-r)=(1-r)(1+r) \\
 & \Rightarrow 3(1-r)=(1+r) \\
\end{align}$
On simplifying,
$\begin{align}
  & \Rightarrow 3-3r=1+r \\
 & \Rightarrow 4r=2 \\
 & \Rightarrow r=\dfrac{1}{2} \\
\end{align}$
Therefore, the first term in the required series is
\[\begin{align}
  & a=3(1-r) \\
 & \text{ }=3(1-\dfrac{1}{2}) \\
 & \text{ }=\dfrac{3}{2} \\
\end{align}\]
So, the terms in the series are \[\dfrac{3}{2},\dfrac{3}{4},\dfrac{3}{8},\dfrac{3}{16},...\]



Option ‘A’ is correct

Note: Here the given sums are in G.P. So, using the sum of infinite terms formula, the required common ratio is evaluated and the series is framed.