Question

Which term of the AP: 3, 15, 27, 39,….. will be 132 more than its 54th term?

Answer 1

Hint: Nth term of an AP can be written using the formula, ${{T}_{n}}=a+\left( n-1 \right)\left( d \right)$

Complete step-by-step answer:

Where a = first term

${{T}_{n}}={{n}^{th}}term$

D = common difference

Solve for ${{T}_{n}}=132+{{T}_{54}}$ and get the value of n. Put the value of ${{T}_{n}}\ and\ {{T}_{54}}$after calculating from the above formula.


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.
Given, AP: 3, 15, 27, 39,…..

Here first term = a = 3.

$\begin{align}

  & {{T}_{1}}=3,{{T}_{2}}=15,{{T}_{3}}=27\ and\ {{T}_{4}}=39..... \\

 & {{T}_{2}}-{{T}_{1}}=15-3=12 \\

 & {{T}_{3}}-{{T}_{2}}=27-15=12 \\

 & {{T}_{4}}-{{T}_{3}}=39-27=12 \\

 & \Rightarrow d=12 \\

\end{align}$

According to the question, we have to find which term is 132 more than 54th term.

Let us assume that the nth term will be 132 more than the 54th term.

i.e. ${{T}_{n}}=132+{{T}_{54}}$……………..(1)

According to formula for ${{T}_{n}}$;

 $\begin{align}

  & {{T}_{n}}=a+\left( n-1 \right)d \\

 & {{T}_{54}}=a+\left( 54-1 \right)d \\

\end{align}$

Putting a = 3 and d = 12, we will get,

$\begin{align}

  & {{T}_{n}}=3+\left( n-1 \right)\times 12 \\

 & {{T}_{n}}=3+12n-12 \\

 & {{T}_{n}}=12n-9 \\

 & and \\

 & {{T}_{54}}=3+\left( 54-1 \right)\times 12 \\

 & {{T}_{54}}=3+53\times 12 \\

 & {{T}_{54}}=3+636 \\

 & {{T}_{54}}=639 \\

\end{align}$

Now, putting these values in equation (1), we will get,

$\Rightarrow 12n-9=132+639$

Taking all the constants in RHS, we will get,

$\begin{align}

  & \Rightarrow 12n=\left( 132+639 \right)+9 \\

 & \Rightarrow 12n=771+9 \\

 & \Rightarrow 12n=780 \\

\end{align}$

Dividing both sides by 12, we will get,

$\begin{align}

  & \Rightarrow n=\dfrac{780}{12} \\

 & \Rightarrow n=65 \\

\end{align}$

Hence, the 65th term of this AP will be 132 more than its 54th term.


Note: Shortcut method:

As, common difference is 12, each term is greater than its previous term by 12. i.e.

Going one step ahead, we are getting a number greater by 12 units.

So, for getting 132 units more, we need to go $\dfrac{132}{12}$ steps ahead.

i.e. 11 steps ahead.

So, the term which will be 132 more than 54th term will be ${{\left( 54+11
\right)}^{th}}$term.

i.e. 65th term.