# Ex.5.2 Q11 Arithmetic Progressions Solution - NCERT Maths Class 10

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## Question

Which term of the AP $$3,\,15,\,27,\,39\dots$$ will be $$132$$ more than its $$54^\rm{th}$$ term?

Video Solution
Arithmetic Progressions
Ex 5.2 | Question 11

## Text Solution

What is Known:?

The AP.

What is Unknown?

Which term will be $$132$$ more than $$54^\rm{th}$$ term.

Reasoning:

$${a_n} = a + \left( {n - 1} \right)d$$ is the general term of AP. Where $${a_n}$$ is the $$n\rm{th}$$ term, $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

Given AP is $$3,\,15,\, 27,\, 39.$$

First term $$a = 3$$

Second term $$a + d = 15$$

$d = 15 - 3 = 12$

$$54^\rm{th}$$ term of the AP is

\begin{align}{a_{54}} &= a + (54 - 1)d\\ &= 3 + 53 \times 12\\ &= 3 + 636\\ &= 639\end{align}

Let $$n^\rm{th}$$ term of AP be $$132$$ more than $$54^\rm{th}$$ term (Given)

We get , $$132\,+\,639=771$$

${a_n} = 771$

\begin{align}{a_n} &= a + \left( {n - 1} \right)d\\771 &= 3 + \left( {n - 1} \right)12\\768 &= \left( {n - 1} \right)12\\\left( {n - 1} \right)& = 64\\n &= 65\end{align}

Therefore, $$65^\rm{th}$$ term will be $$132$$ more than $$54^\rm{th}$$ term.

Alternatively,

Let $$n^\rm{th}$$ term be $$132$$ more than $$54^\rm{th}$$ term.

\begin{align}n &= 54 + \frac{{132}}{{12}}\\& = 54 + 11 = {65^\rm{th}}\,\,{\rm{Term}}\end{align}

$$65^\rm{th}$$ term will be $$132$$ more than $$54^\rm{th}$$ term

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