Question

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

A. 60

B. 62

C. 65

D. 63

Answer 1

Since the given series is an AP so we’ll find the common difference first, then using the formula for the nth term we can find the 54th term. Using the 54th term and the condition given we can find the term which is 132 more than the 54th.

Complete step-by-step solution:

Given the series $3,15,27,39......................$ is in AP.

Let ${a_1} = 3,{a_2} = 15,{a_3} = 27,{a_4} = 39.............$

The common difference of the series is given by ${a_2} - {a_1} = {a_3} - {a_2} = {a_4} - {a_3} = ..................$

Therefore, the common difference of the given series is $d = {a_2} - {a_1} = 15 - 3 = 12$

We know that if a series of $n$ terms is in Arithmetic Progression (A.P) with first term $a$, common difference $d$ then the $nth$ term of the series is given by ${T_n} = a + \left( {n - 1} \right)d$.

In the given problem first term $a = 3$ and common difference $d = 12$

Now we have to find the 54th term i.e., $n = 54$

$\Rightarrow {T_{54}} = 3 + \left( {54 - 1} \right)12$

$\Rightarrow {T_{54}} = 3 + 53\left( {12} \right)$

$\therefore {T_{54}} = 639$

The term we want to find is 132 more than 54th term i.e., 639 + 132 = 771

By substituting ${T_n} = 771$, we can find which term is 771 in the given series of A.P.

$\Rightarrow 771 = 3 + \left( {n - 1} \right)12$

$\Rightarrow 771 - 3 = \left( {n - 1} \right)12$

$\Rightarrow \dfrac{{768}}{{12}} = n - 1$

$\Rightarrow 64 = n - 1$

$\therefore n = 65$

The term which is 132 more than 54th term in the given series of A.P. is 65th term.

So, the correct answer is “Option C”.

Note: An Arithmetic Progression (A.P.) is a sequence of numbers such that the difference of any two successive numbers is a constant. The common difference in an A.P. is given by the difference of the two successive terms in the series.