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

## Question

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

## 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