# Ex.5.3 Q7 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

Find the sum of first \(22\) terms of an AP in which \(d = 7\) and \(22\rm{nd}\) term is \(149.\)

## Text Solution

**What is Known?**

\(d\) and \({a_{22}}\)

**What is Unknown?**

\({S_{22}}\)

**Reasoning:**

Sum of the first \(n\) terms of an AP is given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\) or \({S_n} = \frac{n}{2}\left[ {a + l} \right]\), and \(nth\) term of an AP is\(\,{a_n} = a + \left( {n - 1} \right)d\)

Where \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms and \(l\) is the last term.

**Steps:**

Given,

22nd term, \(l = {a_{22}} = 149\)

Common difference, \(d = 7\)

We know that \(n^\rm{th}\) term of AP, \({a_n} = a + \left( {n - 1} \right)d\)

\[\begin{align}{{a}_{22}}&= a+\left( 22-1 \right)d \\149&=a+21\times 7 \\ 149&=a+147 \\ a &=2 \\\end{align}\]

\[\begin{align}{S_n}&= \frac{n}{2}\left( {a + l} \right)\\ &= \frac{{22}}{2}\left( {2 + 149} \right)\\& = 11 \times 151\\ &= 1661\end{align}\]