Question

How do you find ${{n}^{th}}$ term rule for $\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},\dfrac{1}{16},........?$

Answer 1

Here we have given a sequence $\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},\dfrac{1}{16},.......$. So first of all we will observe that the given sequence is arithmetic progression, geometric progression or arithmetic-geometric progression.
After deciding the given sequence is of which type we will find first term and common ration or common difference accordingly. After that we will use a formula for finding the nth term of the given sequence.
If the sequence is arithmetic sequence than ${{n}^{th}}$ term will be given as $a+\left( n-1 \right)d.$
If the sequence is geometric sequence than ${{n}^{th}}$ term will be given as ${{a}_{n}}=a{{r}^{n-1}}$

Complete Step by Step solution:
The given sequence is $\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},\dfrac{1}{16},........$ ........(i)
Here the common difference is not equal for all the terms. So the given sequence is not arithmetic progression.
Now,
We will find common ratio of given sequence
So, the first terms ${{a}_{1}}=\dfrac{1}{2}$
and $r=$common ration $=\dfrac{\dfrac{1}{4}}{\dfrac{1}{2}}$
$=\dfrac{1\times 2}{4\times 1}$
$r=\dfrac{1}{2}$
So, the given sequence is g.p
So, ${{n}^{th}}$ term is
$\dfrac{1}{2}\times {{\left( \dfrac{1}{2} \right)}^{n-1}}$
$=\dfrac{1}{2}\times \dfrac{1}{{{2}^{n-1}}}$
$=\dfrac{1}{\left( 2 \right)\left( {{2}^{n-1}} \right)}$
$=\dfrac{1}{{{2}^{1+n-1}}}$
$=\dfrac{1}{{{2}^{n}}}$
$\Rightarrow$ The ${{n}^{th}}$ term of

$\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},\dfrac{1}{16},..........=\dfrac{1}{{{2}^{n}}}$.

Note:
Alternative way:
The series is $\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},\dfrac{1}{16},......$ can be written as $\dfrac{1}{{{2}^{1}}},\dfrac{1}{{{2}^{2}}},\dfrac{1}{{{2}^{3}}},\dfrac{1}{{{2}^{4}}},.....$
Hence ${{n}^{th}}$term can be written as $\dfrac{1}{{{2}^{n}}}$.
Geometric series
“Geometric series is also known as geometric progression G.P. is sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-one number called the common ratio given by,
${{s}_{n}}=na$; if $r=1$
Infinite geometric progression (G.P)
Terms of an infinite geometric progression (G.P) can be written as $a,ar,a{{r}^{2}},.....a{{r}^{n-1}},...$ and $a,ar,a{{r}^{2}},....a{{r}^{n-1}},....$ is called infinite geometric series.
The sum of infinite geometric series is given by $\sum\limits_{k=0}^{\infty }{\left( a{{r}^{k}} \right)=a\left( \dfrac{1}{1-r} \right)}$
Arithmetic Progression (AP)
“A mathematical sequence in which the difference between two consecutive terms is always constant and it is abbreviated as arithmetic progression (A.P)”
OR
“An arithmetic sequence or progression is defined as a sequence of numbers in which for every pair of consecutive terms, the second number is obtained by adding a fixed number to the first one.”