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Hint:In this question, we will use ${{n}^{th}}$ term of an AP formula and sum of the first n term of an AP formula to form equations from which we will evaluate values of $a$ and $n$.

__Complete step-by-step answer:__

We know that, an AP with first term $a$ and common difference $d$ is given as,

$a,\,a+d,\,a+2d,\,\ldots \ldots $

Also, we know the formula that, the ${{n}^{th}}$ term of an AP, that is ${{a}_{n}}$, is given by,

${{a}_{n}}=a+\left( n-1 \right)d\cdots \cdots \left( i \right)$

And, the formula to calculate sum of first $n$ terms of an AP, that is ${{S}_{n}}$, is given by,

${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\cdots \cdots \left( ii \right)$

Now, In the given question we have,

${{a}_{n}}=4,\,d=2$

Using these values in equation $\left( i \right)$ , we can write,

$a+\left( n-1 \right)2=4$

Applying distributive law, we get,

$a+2n-2=4$

Adding $2-2n$ on both sides of the equation we get,

$\begin{align}

& a=4+2-2n \\

& \Rightarrow a=6-2n\cdots \cdots \left( iii \right) \\

\end{align}$

Also, we are given that,

${{S}_{n}}=-14$

Using this value in equation $\left( ii \right)$ we have,

$\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)=-14$

Putting value of $d$ and using equation $\left( iii \right)$ here, we get,

$\dfrac{n}{2}\left( 2\left( 6-2n \right)+\left( n-1 \right)2 \right)=-14$

Applying distributive law, we get,

$\begin{align}

& \dfrac{n}{2}\left( 12-4n+2n-2 \right)=-14 \\

& \Rightarrow \dfrac{n}{2}\left( 10-2n \right)=-14 \\

\end{align}$

Applying distributive law again,

$\begin{align}

& \dfrac{10n}{2}-\dfrac{2{{n}^{2}}}{2}=-14 \\

& \Rightarrow 5n-{{n}^{2}}=-14 \\

\end{align}$

Adding, ${{n}^{2}}-5n$ on both sides of the equation, we get,

$\begin{align}

& {{n}^{2}}-5n-14=0 \\

& \Rightarrow {{n}^{2}}-7n+2n-14=0 \\

\end{align}$

Taking common terms, we can write,

$\begin{align}

& n\left( n-7 \right)+2\left( n-7 \right)=0 \\

& \Rightarrow \left( n+2 \right)\left( n-7 \right)=0 \\

\end{align}$

Therefore, either $n+2=0\,$ or $n-7=0$

$\Rightarrow n=-2$ or $n=7$.

But the number of terms cannot be negative,

Hence, $n=7$.

Now, putting $n=7$ in equation $\left( iii \right)$ , we get,

$\begin{align}

& a=6-2n \\

& =6-2\times 7 \\

& =6-14 \\

& =-8 \\

\end{align}$

Hence, the value of $a$ is $-8$ and $n$ is $7$.

Note: In this question, after forming both the equations, we can use any method of solving linear equations in two variables to find the values of $a$ and $n$.Students should remember the general nth term of an AP and sum of first n term of an AP formulas to solve these types of questions.

We know that, an AP with first term $a$ and common difference $d$ is given as,

$a,\,a+d,\,a+2d,\,\ldots \ldots $

Also, we know the formula that, the ${{n}^{th}}$ term of an AP, that is ${{a}_{n}}$, is given by,

${{a}_{n}}=a+\left( n-1 \right)d\cdots \cdots \left( i \right)$

And, the formula to calculate sum of first $n$ terms of an AP, that is ${{S}_{n}}$, is given by,

${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\cdots \cdots \left( ii \right)$

Now, In the given question we have,

${{a}_{n}}=4,\,d=2$

Using these values in equation $\left( i \right)$ , we can write,

$a+\left( n-1 \right)2=4$

Applying distributive law, we get,

$a+2n-2=4$

Adding $2-2n$ on both sides of the equation we get,

$\begin{align}

& a=4+2-2n \\

& \Rightarrow a=6-2n\cdots \cdots \left( iii \right) \\

\end{align}$

Also, we are given that,

${{S}_{n}}=-14$

Using this value in equation $\left( ii \right)$ we have,

$\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)=-14$

Putting value of $d$ and using equation $\left( iii \right)$ here, we get,

$\dfrac{n}{2}\left( 2\left( 6-2n \right)+\left( n-1 \right)2 \right)=-14$

Applying distributive law, we get,

$\begin{align}

& \dfrac{n}{2}\left( 12-4n+2n-2 \right)=-14 \\

& \Rightarrow \dfrac{n}{2}\left( 10-2n \right)=-14 \\

\end{align}$

Applying distributive law again,

$\begin{align}

& \dfrac{10n}{2}-\dfrac{2{{n}^{2}}}{2}=-14 \\

& \Rightarrow 5n-{{n}^{2}}=-14 \\

\end{align}$

Adding, ${{n}^{2}}-5n$ on both sides of the equation, we get,

$\begin{align}

& {{n}^{2}}-5n-14=0 \\

& \Rightarrow {{n}^{2}}-7n+2n-14=0 \\

\end{align}$

Taking common terms, we can write,

$\begin{align}

& n\left( n-7 \right)+2\left( n-7 \right)=0 \\

& \Rightarrow \left( n+2 \right)\left( n-7 \right)=0 \\

\end{align}$

Therefore, either $n+2=0\,$ or $n-7=0$

$\Rightarrow n=-2$ or $n=7$.

But the number of terms cannot be negative,

Hence, $n=7$.

Now, putting $n=7$ in equation $\left( iii \right)$ , we get,

$\begin{align}

& a=6-2n \\

& =6-2\times 7 \\

& =6-14 \\

& =-8 \\

\end{align}$

Hence, the value of $a$ is $-8$ and $n$ is $7$.

Note: In this question, after forming both the equations, we can use any method of solving linear equations in two variables to find the values of $a$ and $n$.Students should remember the general nth term of an AP and sum of first n term of an AP formulas to solve these types of questions.

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