Question

Which term of the A.P. $3,15,27,39,...$ will be $132$ more than its ${54}^{\mathrm{th}}$ term?

Answer 1

Step 1: Determine ${54}^{\mathrm{th}}$ term of the A.P. $3,15,27,39,...$.

Given A.P. is $3,15,27,39,...$.

The first term is $a=3$.

The common difference is:

$$\begin{gathered} d=15-3 \\ \therefore d=12 \end{gathered}$$

The ${54}^{\mathrm{th}}$ term of the given AP, using the $n^{\mathrm{th}}$ term formula, $a_n=a+\left(n-1\right)d$ is:

$\begin{array}{rcl}{a}_{54}& =& 3+\left(54-1\right)12\\ & \Rightarrow & a_{54}=3+53\times 12\\ & \Rightarrow & a_{54}=3+636\\ \therefore a_{54}& =& 639\end{array}$

Step 2. Now, calculate the $n^{\mathrm{th}}$ term of the AP which will be $132$ more than ${54}^{\mathrm{th}}$ term.

So, the required $a_n$ is:

$\begin{array}{rcl}{a}_n& =& 132+639\\ & \Rightarrow & a_n=771\end{array}$

Calculate the $n^{\mathrm{th}}$ term:

$$\begin{gathered} a_n=a+\left(n-1\right)d \\ \Rightarrow 771=3+\left(n-1\right)12 \\ \Rightarrow 771=3+12n-12 \\ \Rightarrow 771=-9+12n \\ \Rightarrow 771+9=12n \\ \Rightarrow 780=12n \\ \Rightarrow 65=n \\ \therefore n=65 \end{gathered}$$

Hence, the ${65}^{\mathrm{th}}$ term will be $132$ more than the ${54}^{\mathrm{th}}$ term.