Question

How do you find the nth term rule for $1,2,4,8,16,32,64$?

Answer 1

Hint:To find the ${n^{th}}$ term, we will have to find which kind of progression is given in the question. So, at first, let us find the common difference between any two consecutive terms in the given progression i.e. $d$.
Here, ${d_1} = 2 - 1 = 1$; ${d_2} = 4 - 2 = 2$ and ${d_3} = 8 - 4 = 4$
So, ${d_1} = 1$, ${d_2} = 2$ and ${d_3} = 4$
Since ${d_1} \ne {d_2} \ne {d_3}$, this implies that the common difference ($d$) is not constant. Therefore the progression is not an arithmetic progression (AP).
Now, let us find the common ratio of the two consecutive terms in the given progression i.e. $r$.
Here, ${r_1} = \dfrac{2}{1} = 2$; ${r_2} = \dfrac{4}{2} = 2$ and ${r_3} = \dfrac{8}{4} = 2$
So, ${r_1} = 2$, ${r_2} = 2$ and ${r_3} = 2$
Since ${r_1} = {r_2} = {r_3}$ , this implies that the common ratio ($r$) is constant and $r = 2$. Therefore the given progression in the question is a geometric progression (GP).

Complete step by step solution:
Given GP is $1,2,4,8,16,32,64$
For a GP, ${n^{th}}$ term is given as
${a_n} = a{r^{n - 1}}$
Where ${a_n} = $${n^{th}}$ term of the GP
$n = $ no. of terms
$a = $ first term
$r = $ common ratio
Here $a = 1$ and $r = 2$. So,
${a_n} = a{r^{n - 1}}$
$
   \Rightarrow {a_n} = 1 \times {2^{n - 1}} \\
   \Rightarrow {a_n} = {2^{n - 1}} \\
 $

Hence, the ${n^{th}}$ term of the sequence $1,2,4,8,16,32,64$ is ${2^{n - 1}}$.

Note: Most of the students get confused between a geometric progression, geometric sequence, and geometric series. In a Geometric sequence, the common ratio of $r$ between successive terms is constant. Progression and sequence mean mostly the same and are used interchangeably. Whereas, a geometric series is the sum of the terms of a geometric sequence.