Question

Which term of the sequence 2, 1, ${{2}^{-1}}$, ${{4}^{-1}}$, ${{8}^{-1}}$,…… is $\dfrac{1}{128}$?
(a) Eighth
(b) Tenth
(c) Ninth
(d) Seventh.

Answer 1

Hint: We start solving the problem by rewriting terms in the sequence using the fact that ${{a}^{-1}}=\dfrac{1}{a}$. We then find the ratios of second and first terms, third term, and second terms which tells us that the given sequence is in geometric series. We then recall the definition of ${{n}^{th}}$ term of geometric progression ${{T}_{n}}=a{{r}^{n-1}}$ to find the required answer.

Complete step-by-step solution:
According to the problem, we have sequence defined as 2, 1, ${{2}^{-1}}$, ${{4}^{-1}}$, ${{8}^{-1}}$,……. We need to find which term is $\dfrac{1}{128}$.
Let us rewrite the given sequence 2, 1, $\dfrac{1}{2}$, $\dfrac{1}{4}$, $\dfrac{1}{8}$,……, as we know ${{a}^{-1}}=\dfrac{1}{a}$.
Let us find the ratio of the second and first term which will be $r=\dfrac{1}{2}$.
Let us find the ratio of third and second term which will be ${{r}_{1}}=\dfrac{\dfrac{1}{2}}{1}=\dfrac{1}{2}$.
We can see that the ratios $r$ and ${{r}_{1}}$ are equal. We know that the ratio between any two consecutive terms in a geometric progression is equal. The given sequence satisfies this which tells us that 2, 1, $\dfrac{1}{2}$, $\dfrac{1}{4}$, $\dfrac{1}{8}$,…… is in geometric progression common ratio $r=\dfrac{1}{2}$.
We know that the equation for the ${{n}^{th}}$ term in the geometric progression is defined as ${{T}_{n}}=a{{r}^{n-1}}$, where a is the first term and r is the common ratio.
From the problem, we have ${{n}^{th}}$ term as $\dfrac{1}{128}$, first term as $2$ and common ratio as $\dfrac{1}{2}$ to find the value of n.
So, we have $2\times {{\left( \dfrac{1}{2} \right)}^{n-1}}=\dfrac{1}{128}$.
$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{n-1}}=\dfrac{1}{128\times 2}$.
$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{n-1}}=\dfrac{1}{256}$.
$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{n-1}}=\dfrac{1}{{{2}^{8}}}$.
$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{n-1}}={{\left( \dfrac{1}{2} \right)}^{8}}$.
We know that if ${{a}^{m}}={{a}^{n}}$, then $m=n$.
$\Rightarrow n-1=8$.
$\Rightarrow n=8+1$.
$\Rightarrow n=9$.
We have found that the ${{9}^{th}}$ term of the sequence is $\dfrac{1}{128}$.
∴ The ${{9}^{th}}$ term of the sequence 2, 1, ${{2}^{-1}}$, ${{4}^{-1}}$, ${{8}^{-1}}$,…… is $\dfrac{1}{128}$.
The correct option for the given problem is (c).

Note: We should not confuse with the formula of ${{n}^{th}}$ term of geometric progression. We should not make calculation mistakes while solving this problem. We can extend this series up to the number of terms we require using the first term and common ratio. We can see that the common ratio is less than 1, using which we can find the sum to the infinity of the series. Similarly, we can expect problems that contain a sequence of terms following Arithmetic or Harmonic progression.