Question

The sum of first n terms of an AP is \[\left( 3{{n}^{2}}+6n \right)\] . Find the ${{n}^{th}}$ term and the ${{15}^{th}}$ term of this AP.

Answer 1

Hint: To solve this question first we will find a and d. Now we know that the sum of n terms of AP is given by $\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ Hence we will rewrite the formula by opening the brackets and compare it with the given sum of n terms which is \[\left( 3{{n}^{2}}+6n \right)\]. On comparing we will get a and d for the given AP. Now we know that ${{n}^{th}}$ term of AP is given by ${{t}_{n}}=a+\left( n-1 \right)d$ . Hence we can easily find the expression for the ${{n}^{th}}$ term. Now substituting n = 15 in this expression we can also find ${{15}^{th}}$ term.

Complete step by step answer:
Now we know that the sum of n terms of AP is given by \[\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] . Where a is the first term of AP and d is the common difference.
Now let's rewrite the formula so that it looks like the given expression.
Hence opening the bracket we get the sum of n terms is $na+\dfrac{n\left( n-1 \right)d}{2}$
$\begin{align}
  & \Rightarrow S =na+\dfrac{d{{n}^{2}}-dn}{2} \\
 & \Rightarrow S =na+\dfrac{d{{n}^{2}}}{2}-\dfrac{dn}{2} \\
 &\Rightarrow S =\dfrac{2na-dn}{2}+\dfrac{d{{n}^{2}}}{2} \\
 &\Rightarrow S =\dfrac{\left( 2a-d \right)n}{2}+\dfrac{d{{n}^{2}}}{2} \\
 &\Rightarrow S =\dfrac{(d){{n}^{2}}}{2}+\dfrac{\left( 2a-d \right)n}{2}.............................\left( 1 \right) \\
\end{align}$
Now we are given that the sum of n terms of the given AP is \[\left( 3{{n}^{2}}+6n \right)\].
Comparing the coefficient of ${{n}^{2}}$ and $n$ from equation (1) we get.
$\dfrac{\left( d \right)}{2}=3$ and $\dfrac{\left( 2a-d \right)}{2}=6$
Hence we have $d=6$ and $2a-d=12$
Now substituting $d=6$ in $2a-d=12$ We get
$\begin{align}
  & 2a-6=12 \\
 & \Rightarrow 2a=12+6 \\
 & \Rightarrow 2a=18 \\
 & \therefore a=9 \\
\end{align}$
Hence we have a = 9 and d = 6.
Now let us find ${{n}^{th}}$ term of this AP.
Now we know for an AP with first term as a and common difference as d the ${{n}^{th}}$ term is given by $a+\left( n-1 \right)d$ .
Now substituting a = 9 and d = - 6 we get.
${{n}^{th}}$ term of AP is $9+\left( n-1 \right)\left( 6 \right)$
Hence we have ${{n}^{th}}$ term is
$\begin{align}
  & {{t}_{n}}=9+6n-6 \\
 & \therefore {{t}_{n}}=6n+3 \\
\end{align}$
Hence we have ${{n}^{th}}$ term of the AP is ${{t}_{n}}=6n+3$
Now substituting n = 15 we get
${{t}_{15}}=6\left( 15 \right)+3=90+3=93$

Hence the ${{15}^{th}}$ term of AP is 93.

Note: Now note that the formula for ${{n}^{th}}$ term of an AP is given by ${{t}_{n}}=a+\left( n-1 \right)d$ and the sum of ${{n}^{th}}$ terms of AP is given by $\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ . Also since we are given with the expression of the sum of n terms we compare the expression with our formula. In this solution we have tried to convert the formula to look like a given expression. Though we can also convert the given expression to look like the formula $\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ . Though in both cases we will compare and find a and d.