Question

If the geometric expression 162, 54, 18……….. and $\dfrac{2}{81},\dfrac{2}{27},\dfrac{2}{9},..........$ have their $n^{th}$ term equal. Find the value of n.

Answer 1

Hint: Write individually, the $n^{th}$ term of both the geometric expressions and then equate both the $n^{th}$ terms and then find the value of n at which both the $n^{th}$ terms of the geometric expression are equal.

Complete step-by-step solution -
The geometric expressions given in the question are:
162, 54, 18………..
 $\dfrac{2}{81},\dfrac{2}{27},\dfrac{2}{9},..........$
We know the $n^{th}$ term of a geometric progression is:
${{T}_{n}}=a{{r}^{n-1}}$
In the above equation, “a” is the first term and “r” is the common ratio of the given geometric progression.
We are going to write the nth terms for the following geometric progression.
162, 54, 18………..
In the above series, first term of the series is $a=162$ and common ratio is $r=\dfrac{1}{3}$.
Writing the $n^{th}$ term of the above geometric progression by plugging the values of $a=162$ and $r=\dfrac{1}{3}$ we get,
$\begin{align}
  & {{T}_{n}}=a{{r}^{n-1}} \\
 & \Rightarrow {{T}_{n}}=162{{\left( \dfrac{1}{3} \right)}^{n-1}} \\
 & \Rightarrow {{T}_{n}}=\dfrac{162}{{{3}^{n-1}}} \\
\end{align}$
We are going to write the $n^{th}$ terms for the following geometric progression.
$\dfrac{2}{81},\dfrac{2}{27},\dfrac{2}{9},..........$
In the above series, first term of the series is $a=\dfrac{2}{81}$ and common ratio is $r=3$.
Writing the nth term of the above geometric progression by plugging the values of $a=\dfrac{2}{81}$ and $r=3$ we get,
$\begin{align}
  & {{T}_{n}}=a{{r}^{n-1}} \\
 & \Rightarrow {{T}_{n}}=\dfrac{2}{81}{{\left( 3 \right)}^{n-1}} \\
 & \Rightarrow {{T}_{n}}=\dfrac{2}{{{3}^{4}}}{{\left( 3 \right)}^{n-1}} \\
 & \Rightarrow {{T}_{n}}=2{{\left( 3 \right)}^{n-5}} \\
\end{align}$
Now, on equating the nth term of both the geometric progressions we get,
$\dfrac{162}{{{3}^{n-1}}}=2{{\left( 3 \right)}^{n-5}}$
On cross-multiplication of above relation we get,
$162=2{{\left( 3 \right)}^{n-5}}{{\left( 3 \right)}^{n-1}}$
There is a property that if the base is same then powers of the bases are added.
$\begin{align}
  & 162=2{{\left( 3 \right)}^{2n-6}} \\
 & \Rightarrow 81={{\left( 3 \right)}^{2n-6}} \\
 & \Rightarrow {{3}^{4}}={{\left( 3 \right)}^{2n-6}} \\
 & \Rightarrow 4=2n-6 \\
\end{align}$
$\begin{align}
  & \Rightarrow 10=2n \\
 & \Rightarrow n=5 \\
\end{align}$
From the above solution, we have got the value of n equal to 5.
Hence, the value of n is equal to 5.

Note: You might want to know how we have got the value of the common ratio in the above solution.
We are showing the procedure of finding the common ratio of the following geometric progression.
162, 54, 18………..
The common ratio is the division of any term of a geometric progression by its preceding term.
Let us take any term from the geometric progression as 18, now we are going to divide 18 by its preceding term i.e. 54 so the common ratio will look as below:
$\dfrac{18}{54}=\dfrac{1}{3}$
Similarly, you can check this value of the common ratio in the given geometric progression by taking the term different from that we have shown above.