# Ex.5.2 Q8 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

An AP consists of \(50\) terms of which \(3^\rm{rd}\) term is \(12\) and the last term is \(106.\) Find the \(29^\rm{th}\) term.

## Text Solution

**What is Known:?**

Number of terms in the AP**. **\(3^\rm{rd}\) term is \(12\) and last term is \(106\)

**What is Unknown?**

\(29^\rm{th}\) 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:**

\[{a_n} = a + \left( {n - 1} \right)d\]

Third term of AP is \(a + 2d\)

\[a + 2d = 12 \qquad \dots(1)\]

Last term \(= 106\)

\(50\rm{th}\) term \(=106\)

\[\begin{align}a + \left( {50 - 1} \right)d &= 106\\a + 49d &= 106 \qquad \dots\left( 2 \right)\end{align}\]

Solving (1) & (2) for the values of \(a\) and \(d\).

\[\begin{align}47d &= 94\\d &= 2\end{align}\]

Putting \(d = 2\) in equation (1)

\[\begin{align}a + 2 \times 2 &= 12\\a + 4 &= 12\\a &= 12 - 4\\a &= 8\end{align}\]

\(29^\rm{th}\) term of AP is

\[\begin{align}{a_{29}} &= a + (29 - 1)d\\{a_{29}} &= 8 + \left( {28} \right)2\\{a_{29}} &= 8 + 56\\{a_{29}}& = 64\end{align}\]

\(29^\rm{th}\) term of AP is \(64.\)