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**Hint:**To solve this question, firstly we will find the values of ${{S}_{1}}$ and ${{S}_{2}}$ by using relation given in question, which is ${{S}_{n}}=3{{n}^{2}}-4n$. After that we will find the values of first two terms of an A.P by solving the values obtained of ${{S}_{1}}$ and ${{S}_{2}}$ as ${{S}_{1}}={{a}_{1}}$ and ${{S}_{2}}={{a}_{1}}+{{a}_{2}}$. Then, we will find the common difference d, and then, we will substitute all the values in the formula of ${{n}^{th}}$ term of an A.P, to get the value of ${{n}^{th}}$ term.

**Complete step-by-step solution**Let, terms of A.P be ${{a}_{1}},{{a}_{2}},{{a}_{3}},.......,{{a}_{n}}$

Now, in question it is given that the sum of first n terms of an A.P. is given by

${{S}_{n}}=3{{n}^{2}}-4n$

So, if we substitute $n = 1$, we will get the value of ${{S}_{1}}$, which will be equals to the first term of an A.P which is ${{a}_{1}}$.

Then, ${{S}_{1}}=3{{(1)}^{2}}-4(1)$

${{S}_{1}}=-1$

Substituting n = 2, we will get value of ${{S}_{2}}$, which will be equals to sum of first two terms which is ${{a}_{1}}+{{a}_{2}}$

Then, ${{S}_{2}}=3{{(2)}^{2}}-4(2)$

${{S}_{1}}=4$

Also, we know that, ${{S}_{n}}-{{S}_{n-1}}={{a}_{n}}$

So, ${{S}_{2}}-{{S}_{1}}={{a}_{2}}$

$\Rightarrow {{a}_{2}}=4-(-1)$

$\Rightarrow {{a}_{2}}=5$

So, A.P will be $-1, 5, ………$

Then, first term a = -1 and common difference $d = 5 – (-1) = 6$

So, we know that ${{n}^{th}}$ term of an A.P is given by formula,

${{a}_{n}}=a+(n-1)d$

Putting values of a = -1 and d = 6, we get

${{a}_{n}}=-1+(n-1)6$

On simplifying, we get

${{a}_{n}}=-1+6n-6$

On solving, we get

${{a}_{n}}=6n-7$

**Hence, ${{n}^{th}}$ term of an A.P is equal to $6n – 7.$**

**Note:**Always remember that ${{S}_{n}}-{{S}_{n-1}}={{a}_{n}}$, ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$ and ${{n}^{th}}$ term of an A.P is given by formula,${{a}_{n}}=a+(n-1)d$, where d is common difference between consecutive terms and a is first term of an A.P. While solving the question, always take care of signs of the term and try not to make any calculation mistakes.

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