# Ex.5.3 Q5 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

The first term of an AP is \(5,\) the last term is \(45\) and the sum is \(400.\) Find the number of terms and the common difference.

## Text Solution

**What is Known?**

\(a,{\rm{ }}l\), and \({S_n}\)

**What is Unknown?**

\(n\) and \(l\)

**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,

- First term, \(a = 5\)
- Last term, \(l = 45\)
- Sum up to \(n\rm{th}\) terms, \({S_n} = 400\)

We know that sum of \(n\) terms of AP

\[\begin{align}{S_n} &= \frac{n}{2}\left( {a + l} \right)\\400& = \frac{n}{2}\left( {5 + 45} \right)\\400&= \frac{n}{2} \times 50\\n &= 16\end{align}\]

\[\begin{align}l &= {a_n} = a + \left( {n - 1} \right)d\\45 &= 5 + \left( {16 - 1} \right)d\\40 &= 15d\\d &= \frac{{40}}{{15}}\\d& = \frac{8}{3}\end{align}\]