Question

Let \[{{A}_{n}}\] be the sum of the first $n$ terms of the geometric series \[704+\dfrac{704}{2}+\dfrac{704}{4}+\dfrac{704}{8}+...\] and \[{{B}_{n}}\] be the sum of the first $n$ terms of the geometric series \[1984+\dfrac{1984}{2}+\dfrac{1984}{4}+\dfrac{1984}{8}+...\]. If \[{{A}_{n}}={{B}_{n}}\], then the value of $n$ is (where \[n\in N\])
A. \[4\]
B. \[5\]
C. \[6\]
D. \[7\]

Answer 1

Hint: In this question, we have to find the number of terms in the given series. For this, the sum of $n$ terms of both the series to be found and by substituting those values in the given condition \[{{A}_{n}}={{B}_{n}}\]. On simplifying, we get the required $n$ terms.

Formula Used: If the series is a geometric series, then the sum of the $n$ terms of a geometric series is calculated by
 ${{S}_{n}}=\dfrac{a({{r}^{n}}-1)}{r-1}$ where $r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{n}}$- Sum of the $n$ terms; $n$ - Number of terms; $a$ - First term; $r$- Common ratio.

Complete step by step solution: Given geometric series are:
\[{{A}_{n}}=704+\dfrac{704}{2}+\dfrac{704}{4}+\dfrac{704}{8}+...\]
And
\[{{B}_{n}}=1984+\dfrac{1984}{2}+\dfrac{1984}{4}+\dfrac{1984}{8}+...\]
Then, the sum of the $n$ terms of the series \[{{A}_{n}}\] is
\[\begin{align}
  & a=704;r=\dfrac{{}^{704}/{}_{2}}{704}=\dfrac{1}{2} \\
 & {{A}_{n}}=\dfrac{704\left( 1-{{\left( \dfrac{1}{2} \right)}^{n}} \right)}{1-\dfrac{1}{2}} \\
 & \text{ }=704\times 2\left( 1-{{\left( \dfrac{1}{2} \right)}^{n}} \right) \\
\end{align}\]
The sum of the $n$ terms of the series \[{{B}_{n}}\] is
\[\begin{align}
  & a=1984;r=\dfrac{{}^{1984}/{}_{2}}{1984}=\dfrac{1}{2} \\
 & {{B}_{n}}=\dfrac{1984\left( 1-{{\left( \dfrac{1}{2} \right)}^{n}} \right)}{1-\dfrac{1}{2}} \\
 & \text{ }=1984\times \dfrac{2}{3}\left( 1-{{\left( \dfrac{-1}{2} \right)}^{n}} \right) \\
\end{align}\]
But it is given that, \[{{A}_{n}}={{B}_{n}}\]
So, on equating, we get
\[\begin{align}
  & {{A}_{n}}={{B}_{n}} \\
 & \Rightarrow 704\times 2\left( 1-{{\left( \dfrac{1}{2} \right)}^{n}} \right)=1984\times \dfrac{2}{3}\left( 1-{{\left( \dfrac{-1}{2} \right)}^{n}} \right) \\
 & \Rightarrow 2112\left( 1-{{\left( \dfrac{1}{2} \right)}^{n}} \right)=1984\left( 1-{{\left( \dfrac{-1}{2} \right)}^{n}} \right) \\
 & \Rightarrow 33\left( 1-{{\left( \dfrac{1}{2} \right)}^{n}} \right)=31\left( 1-{{\left( \dfrac{-1}{2} \right)}^{n}} \right) \\
 & \Rightarrow 33-33{{\left( \dfrac{1}{2} \right)}^{n}}=31-31{{\left( \dfrac{-1}{2} \right)}^{n}} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow 33-31=33{{\left( \dfrac{1}{2} \right)}^{n}}-31{{\left( \dfrac{-1}{2} \right)}^{n}} \\
 & \Rightarrow 2\times {{2}^{n}}=33{{(1)}^{n}}-31{{(-1)}^{n}} \\
 & \Rightarrow {{2}^{n+1}}=33-31{{(-1)}^{n}} \\
 & \Rightarrow {{2}^{n+1}}+31{{(-1)}^{n}}=33\text{ }...(1) \\
\end{align}\]
Then, substituting the given options for $n$ in (1), we get
If \[n=4;\] then
\[\begin{align}
  & {{2}^{n+1}}+31{{(-1)}^{n}}={{2}^{4+1}}+31{{(-1)}^{4}} \\
 & \text{ }={{2}^{5}}+31 \\
 & \text{ =}32+31 \\
 & \text{ =}63\ne 33 \\
\end{align}\]
Thus, the first option is incorrect.
If \[n=5;\] then
\[\begin{align}
  & {{2}^{n+1}}+31{{(-1)}^{n}}={{2}^{5+1}}+31{{(-1)}^{5}} \\
 & \text{ }={{2}^{6}}-31 \\
 & \text{ }=64-31=33 \\
\end{align}\]
Thus, the required value of the number of terms in the given series is \[n=5\].

Option ‘B’ is correct

Note: Here, we have to substitute the given values of $n$ in the obtained equation, in order to get the correct value. Since the given series are in geometric progression, we can use the sum of $n$ terms formula for finding the required value.