Question

The ${{n}^{th}}$ term of the series 16, 8, 4, .…… is $\dfrac{1}{{{2}^{17}}}$. Find the value of n?

Answer 1

If you carefully look at the pattern of the series, you will find that the series is in G.P. (Geometric Progression) with first term as $\left( a=16 \right)$ and common ratio as $\left( r=\dfrac{1}{2} \right)$. We know that the general term of the G.P. is as follows: ${{T}_{n}}=a{{r}^{n-1}}$. So, using this general term formula, we can find the value of “n”.

Complete step by step answer:
The series given in the above problem is as follows:
16, 8, 4,………..
As you can see that the above series is in G.P. with the first term $\left( a=16 \right)$ and common ratio as $\left( r=\dfrac{1}{2} \right)$.
A common ratio is calculated by taking any term and then divides this term to its successive term. Let us do this common ratio calculation in the above problem by taking any term say 8 and the successive term from 8 is 4 so dividing 4 by 8 we get,
$\begin{aligned} & \dfrac{4}{8} \\\ & =\dfrac{1}{2} \\\ \end{aligned}$
Now, we are going to use the general term of the G.P. which is in the following form:
${{T}_{n}}=a{{r}^{n-1}}$
In the above formula, $''a''$ is the first term, “r” is the common ratio and “n” is the order of the term.
Substituting the value of $\left( a=16 \right)$ and $\left( r=\dfrac{1}{2} \right)$ in the above equation we get,
${{T}_{n}}=\left( 16 \right){{\left( \dfrac{1}{2} \right)}^{n-1}}$
In the above problem, we have given the value of ${{n}^{th}}$ term as $\dfrac{1}{{{2}^{17}}}$ so substituting the value of ${{T}_{n}}=\dfrac{1}{{{2}^{17}}}$ in the above equation we get,
$\begin{aligned} & \dfrac{1}{{{2}^{17}}}=\left( 16 \right){{\left( \dfrac{1}{2} \right)}^{n-1}} \\\ & \Rightarrow \dfrac{1}{{{2}^{17}}}=\left( 16 \right)\left( \dfrac{1}{{{2}^{n-1}}} \right) \\\ \end{aligned}$
We can write 16 as 2 to the power of 4 in the above equation and we get,
$\begin{aligned} & \Rightarrow \dfrac{1}{{{2}^{17}}}=\left( {{2}^{4}} \right)\left( \dfrac{1}{{{2}^{n-1}}} \right) \\\ & \Rightarrow \dfrac{1}{{{2}^{17}}}=\dfrac{{{2}^{4}}}{{{2}^{n-1}}} \\\ \end{aligned}$
On cross-multiplying the above equation we get,
${{2}^{n-1}}={{2}^{4}}\left( {{2}^{17}} \right)$
There is a property which states that if the base is same and are written with the multiplication sign then we can add the exponents of the bases so in the R.H.S of the above equation you can see base 2 is same with different exponents and the base 2 is written with a multiplication sign then we get,
$\begin{aligned} & {{2}^{n-1}}={{2}^{4+17}} \\\ & \Rightarrow {{2}^{n-1}}={{2}^{21}} \\\ \end{aligned}$
In the above equation, base 2 is same on both the sides so we can equate their powers and we get,
$\begin{aligned} & n-1=21 \\\ & \Rightarrow n=22 \\\ \end{aligned}$
From the above solution, we got the value of “n” as 22.

Note: The mistake that could be possible in the above problem is that while finding the common ratio you might divide the larger number by a smaller number because this is the kind of problem we generally see in which we are dividing higher numbers by lower numbers. But in this problem, we have a decreasing G.P. so we have to divide a lower number by a higher number so make sure you won’t make this mistake.