Given in an A.P. ${{a}_{n}}=4,d=2,{{S}_{n}}=-14,$ find n and a.

Hint: In question, standard notations are used. ‘a’ is used for the first term of an AP, $'{{a}_{n}}'$ is used for nth term of the A.P., ‘d’ is used for the common difference of the A.P. and $'{{S}_{n}}'$is used for sum of all the terms of the A.P. up to nth term,

Use the formula $''{{T}_{n}}=a+\left( n-1 \right)d''\ and\ ''{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]''$to get the values of ‘d’ and ‘n’.

Complete step-by-step answer:

A series is called arithmetic progression (AP) if the difference between any of its consecutive terms is the same.

Common difference (d) of an AP is defined as the difference between its consecutive terms.
Here ${{a}_{n}}=4,d=2,{{S}_{n}}=-14,$

Let us assume the first term to be ‘a’.

Given, ${{a}_{n}}=4$

i.e. our nth term is 4.

We know, ${{T}_{n}}=a+\left( n-1 \right)d.$

Putting values, we will get;

$\begin{align}

  & \Rightarrow 4=a+\left( n-1 \right)\times 2 \\

 & \Rightarrow 4=a+2n-2 \\

 & \Rightarrow 4=\left( a-2 \right)+2n \\

\end{align}$

Taking term of ‘n’ to LHS and all constant terms to RHS, we will get;

$\begin{align}

  & \Rightarrow -2n=a-2-4 \\

 & \Rightarrow -2n=a-6 \\

 & \Rightarrow a=-2n+6.................\left( 1 \right) \\

\end{align}$

We know ,${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$

Putting ${{S}_{n}}=-14\ and\ d=2$, we will get;

$\begin{align}

  & \Rightarrow -14=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)\times 2 \right] \\

 & \Rightarrow -14=\dfrac{n}{2}\left[ 2a+2n-2 \right] \\

\end{align}$

Taking $\dfrac{1}{2}$to inside of the bracket in RHS;

$\begin{align}

  & \Rightarrow -14=n\left[ \dfrac{2a+2n-2}{2} \right] \\

 & \Rightarrow -14=n\left( a+n-1 \right) \\

\end{align}$

Putting value of ‘a’ from equation (1), we will get;

\[\begin{align}

  & \Rightarrow -14=n\times \left[ \left( -2n+6 \right)+n-1 \right] \\

 & \Rightarrow -14=n\times \left[ -n+5 \right] \\

 & \Rightarrow -14=-{{n}^{2}}+5n \\

\end{align}\]

Taking all the terms to one side of equality sign;

\[\Rightarrow {{n}^{2}}-5n-14=0\]

Now, we have got a quadratic equation in n for showing this equation, let’s factorize the “quadratic part”.

For factoring the quadratic polynomial, let’s split middle term and factorise;

\[\begin{align}

  & \Rightarrow {{n}^{2}}-5n-14=0 \\

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

\end{align}\]

Taking ‘n’ common from first two terms and ‘2’ common from last two terms, we will get;

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

Taking \[\left( n-7 \right)\] common, we will get;

\[\begin{align}

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

 & \Rightarrow n-7=0\ and/or\ n+2=0 \\

 & \Rightarrow n=7\ and\ n=-2 \\

\end{align}\]

 ‘n’ is the number of terms, so it can’t be negative, so, required n = 7.

Now, we have to find a;

To find ‘a’ let’s put values of ‘n’ in equation (1);

$\begin{align}

  & a=-2n+6 \\

 & a=\left( -2 \right)\left( 7 \right)+6 \\

 & a=-14+6 \\

 & a=-8 \\

\end{align}$

Hence, required n = 7 and a = -8.


Note: On solving the obtained quadratic equation, we have got two values of n i.e. n = 7, and n = -2. But as ‘n’ is used for the nth term, n can’t be negative so, n = is the only value which is acceptable.