Question

If $S, P$, and $R$ are the sum, product, and sum of the reciprocals of the $n$ terms of an increasing GP respectively and ${{S}^{n}}={{R}^{n}}\cdot {{P}^{k}}$, then $k$ is equal to
A. $1$
B. $2$
C. $3$
D. None of these

Answer 1

Hint: In this question, we have to find the value of the given variable $k$ for the given equation ${{S}^{n}}={{R}^{n}}\cdot {{P}^{k}}$. Since it is given that $S, P$, and $R$ are the sum, product, and sum of the reciprocals of the $n$ terms of an increasing GP, by using appropriate formulae, we can find the required value.

Formula Used: The sum of $n$ terms of a geometric series is
${{S}_{n}}=\dfrac{a(1-{{r}^{n}})}{1-r}$ where $r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{n}}$ - Sum of $n$ terms of the series; $n$ - Number of terms; $r$ - Common ratio; $a$ - First term
The product of the $n$ term of a geometric series is
${{P}_{n}}={{a}^{n}}\cdot {{r}^{\left( \dfrac{n(n-1)}{2} \right)}}$

Complete step by step solution: Given that, $S, P$ and $R$ are the sum, product, and sum of the reciprocals of the $n$ terms of an increasing GP respectively.
So, we can write
The sum $S=\dfrac{a(1-{{r}^{n}})}{1-r}$;
The product $P={{a}^{n}}\cdot {{r}^{\left( \dfrac{n(n-1)}{2} \right)}}$;
And
The sum of the reciprocal of $n$ terms of an increasing GP is
$\begin{align}
  & R=\dfrac{1}{a}+\dfrac{1}{ar}+\dfrac{1}{a{{r}^{2}}}+....\dfrac{1}{a{{r}^{n}}} \\
 & \text{ }=\dfrac{\dfrac{1}{a}\left( 1-{{\left( \dfrac{1}{r} \right)}^{n}} \right)}{1-\dfrac{1}{r}} \\
\end{align}$
$\begin{align}
  & \text{ }=\dfrac{1-\dfrac{1}{{{r}^{n}}}}{a\left( 1-\dfrac{1}{r} \right)} \\
 & \text{ }=\dfrac{1-{{r}^{n}}}{a\left( 1-r \right){{r}^{n-1}}} \\
\end{align}$
But we have ${{S}^{n}}={{R}^{n}}\cdot {{P}^{k}}$
On simplifying,
$\begin{align}
  & {{S}^{n}}={{R}^{n}}\cdot {{P}^{k}} \\
 & \Rightarrow \dfrac{{{S}^{n}}}{{{R}^{n}}}={{P}^{k}} \\
 & \Rightarrow {{\left( \dfrac{S}{R} \right)}^{n}}={{P}^{k}} \\
\end{align}$
So, substituting the values of $S,P$ and $R$ in the above expression, we get
$\begin{align}
  & {{\left( \dfrac{S}{R} \right)}^{n}}={{P}^{k}} \\
 & \Rightarrow {{\left( {{a}^{2}}{{r}^{n-1}} \right)}^{n}}={{P}^{k}}\text{ }...(1) \\
\end{align}$
But we have $P={{a}^{n}}\cdot {{r}^{\left( \dfrac{n(n-1)}{2} \right)}}$
We can write it by squaring on both sides as
$\begin{align}
  & {{P}^{2}}={{\left( {{a}^{n}}\cdot {{r}^{\left( \dfrac{n(n-1)}{2} \right)}} \right)}^{2}} \\
 & \text{ }={{\left( {{a}^{2}}\cdot {{r}^{\left( \dfrac{2(n-1)}{2} \right)}} \right)}^{n}} \\
 & \text{ }={{\left( {{a}^{2}}\cdot {{r}^{n-1}} \right)}^{n}} \\
\end{align}$
So, substituting in (1), we get
$\begin{align}
  & {{\left( {{a}^{2}}{{r}^{n-1}} \right)}^{n}}={{P}^{k}} \\
 & \Rightarrow {{P}^{2}}={{P}^{k}} \\
 & \therefore k=2 \\
\end{align}$


Option ‘B’ is correct

Note: Here we need to apply the appropriate formula for the given sum, product, and sun of the reciprocals of a geometric series. So, that we can find the required value by substituting them in the given expression.