For a sequence if $ {{S}_{n}}=\dfrac{{{4}^{n}}-{{3}^{n}}}{{{3}^{n}}} $ find the $ {{n}^{th}} $ term, hence show that if it is a G.P.

Hint: In the problem, we have the value of sum of the $ n $ terms in a series. Now we will simplify the given value and calculate the value of the sum of $ n-1 $ terms from the value of $ {{S}_{n}} $. The value of $ {{n}^{th}} $ can be calculated by subtracting the sum of $ n-1 $ terms from the sum of $ n $ terms. After finding the $ {{n}^{th}} $ term, we will calculate the first, second, third terms of the series by using the value of $ {{n}^{th}} $ term. We will calculate the ratio of first, second term, and the ratio of a second and third term. From these ratios, we can say that the given series is in G.P or not.

Complete step by step answer:
Given that,
 $ {{S}_{n}}=\dfrac{{{4}^{n}}-{{3}^{n}}}{{{3}^{n}}} $
Dividing each term in numerator with the denominator, then we will get
 $ \Rightarrow {{S}_{n}}=\dfrac{{{4}^{n}}}{{{3}^{n}}}-\dfrac{{{3}^{n}}}{{{3}^{n}}} $
We know that $ \left( \dfrac{{{a}^{m}}}{{{b}^{m}}} \right)={{\left( \dfrac{a}{b} \right)}^{m}} $ , then we will have
 $ \Rightarrow {{S}_{n}}={{\left( \dfrac{4}{3} \right)}^{n}}-1 $
From the value of $ {{S}_{n}} $ , the sum of $ n-1 $ terms can be given as
 $ {{S}_{n-1}}={{\left( \dfrac{4}{3} \right)}^{n-1}}-1 $
We know that the $ {{n}^{th}} $ term is equal to the difference between the sum of $ n $ terms and the sum of $ n-1 $ terms. Mathematically
 $ \begin{align}
  & {{T}_{n}}={{S}_{n}}-{{S}_{n-1}} \\
 & \Rightarrow {{T}_{n}}=\left[ {{\left( \dfrac{4}{3} \right)}^{n}}-1 \right]-\left[ {{\left( \dfrac{4}{3} \right)}^{n-1}}-1 \right] \\
 & \Rightarrow {{T}_{n}}={{\left( \dfrac{4}{3} \right)}^{n}}-1-{{\left( \dfrac{4}{3} \right)}^{n-1}}+1 \\
\end{align} $
We can write $ {{a}^{m}}={{a}^{m-1}}.a $ , then we will get
 $ {{T}_{n}}={{\left( \dfrac{4}{3} \right)}^{n-1}}.\left( \dfrac{4}{3} \right)-{{\left( \dfrac{4}{3} \right)}^{n-1}} $
Taking $ {{\left( \dfrac{4}{3} \right)}^{n-1}} $ as common in the above equation, then we will get
 $ \begin{align}
  & {{T}_{n}}={{\left( \dfrac{4}{3} \right)}^{n-1}}\left[ \dfrac{4}{3}-1 \right] \\
 & \Rightarrow {{T}_{n}}={{\left( \dfrac{4}{3} \right)}^{n-1}}\left[ \dfrac{4-3}{3} \right] \\
 & \Rightarrow {{T}_{n}}=\dfrac{1}{3}{{\left( \dfrac{4}{3} \right)}^{n-1}} \\
\end{align} $
Multiplying and dividing with $ 4 $ , then we will get
 $ \begin{align}
  & {{T}_{n}}=\dfrac{1}{4}.\dfrac{4}{3}{{\left( \dfrac{4}{3} \right)}^{n-1}} \\
 & \Rightarrow {{T}_{n}}=\dfrac{1}{4}{{\left( \dfrac{4}{3} \right)}^{n-1+1}} \\
 & \Rightarrow {{T}_{n}}=\dfrac{1}{4}{{\left( \dfrac{4}{3} \right)}^{n}} \\
\end{align} $
Now we have the value of $ {{n}^{th}} $ term of the given series. Now the first, second, third terms of the series are
First term is $ {{T}_{1}}=\dfrac{1}{4}{{\left( \dfrac{4}{3} \right)}^{1}}=\dfrac{1}{3} $ .
Second term is $ {{T}_{2}}=\dfrac{1}{4}{{\left( \dfrac{4}{3} \right)}^{2}}=\dfrac{4}{9} $ .
Third term is $ {{T}_{3}}=\dfrac{1}{4}{{\left( \dfrac{4}{3} \right)}^{3}}=\dfrac{16}{27} $ .
Now the ratio of second and first term is $ \dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{\dfrac{4}{9}}{\dfrac{1}{3}}=\dfrac{4}{3} $
Now the ratio of third and second term is $ \dfrac{{{T}_{3}}}{{{T}_{2}}}=\dfrac{\dfrac{16}{27}}{\dfrac{4}{9}}=\dfrac{4}{3} $
Here we got the ratio of successive terms as equal. When the ratio of the successive terms is equal in a series, then we will call that series as G.P.
So, the given series is in G.P with common ratio $ r=\dfrac{4}{3} $ .

Note:
We can also solve this problem in another manner. Once we calculated the $ {{n}^{th}} $ term, to show the series in G.P we will modify the given $ {{S}_{n}} $ value and convert it in the form such that the sum of $ n $ terms in G.P i.e. $ S=\dfrac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)} $ .
Given $ {{S}_{n}}=\dfrac{{{4}^{n}}-{{3}^{n}}}{{{3}^{n}}} $
 $ \Rightarrow {{S}_{n}}=\dfrac{{{4}^{n}}}{{{3}^{n}}}-\dfrac{{{3}^{n}}}{{{3}^{n}}} $
 $ \Rightarrow {{S}_{n}}={{\left( \dfrac{4}{3} \right)}^{n}}-1 $
Dividing and multiplying with $ 3 $ , then we will get\
 $ \begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{3}{3}{{\left( \dfrac{4}{3} \right)}^{n}}-1 \\
 & \Rightarrow {{S}_{n}}=\dfrac{\dfrac{1}{3}\left[ {{\left( \dfrac{4}{3} \right)}^{n}}-1 \right]}{\dfrac{1}{3}} \\
\end{align} $
Adding and subtracting $ 1 $ in denominator, then we will get
 $ \begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{\dfrac{1}{3}\left[ {{\left( \dfrac{4}{3} \right)}^{n}}-1 \right]}{\dfrac{1}{3}+1-1} \\
 & \Rightarrow {{S}_{n}}=\dfrac{\dfrac{1}{3}\left[ {{\left( \dfrac{4}{3} \right)}^{n}}-1 \right]}{\dfrac{1+1\times 3}{3}-1} \\
 & \Rightarrow {{S}_{n}}=\dfrac{\dfrac{1}{3}\left[ {{\left( \dfrac{4}{3} \right)}^{n}}-1 \right]}{\left( \dfrac{4}{3}-1 \right)} \\
\end{align} $
The above is in the form of $ S=\dfrac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)} $ with $ r=\dfrac{4}{3} $ .
Hence the series in G.P.