Question

In an AP given ${{a}_{n}}=4,\,d=2,\,{{S}_{n}}=-14$, find $n$ and $a$.

Answer 1

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.