# Ex.5.3 Q6 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

The first and the last term of an AP are \(17\) and \(350\) respectively. If the common difference is \(9,\) how many terms are there and what is their sum?

## Text Solution

**What is known?**

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

**What is unknown?**

\(n\) and \({S_n}\)

**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 \(n \rm{th}\) 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, \(a = 17\)
- Last term, \(l = 350\)
- Common difference, \(d = 9\)

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

\[\begin{align}350& = 17 + \left( {n - 1} \right)9\\333 &= \left( {n - 1} \right)9\\\left( {n - 1} \right) &= 37\\n &= 38\end{align}\]

Sum of \(n\) terms of AP,

\[\begin{align}{S_n} &= \frac{n}{2}\left( {a + l} \right)\\{S_{38}} &= \frac{{38}}{2}\left( {17 + 350} \right)\\

\, &= 19 \times 367\\ &= 6973\end{align}\]

Thus, this A.P. contains \(38\) terms and the sum of the terms of this A.P. is \(6973.\)