# Ex.5.3 Q8 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

Find the sum of first \(51\) terms of an AP whose second and third terms are \(14\) and \(18\) respectively.

## Text Solution

**What is known?**

\({a_2}\) and \({a_3}\)

**What is unknown?**

\({S_{51}}\)

**Reasoning:**

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

2nd term, \({a_2} = 14\)

3rd term, \({a_3} = 18\)

Common difference, \(d = {a_3} - {a_2} = 18 - 14 = 4\)

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

\[\begin{align}{a_2} &= a + d\\14 &= a + 4\\a& = 10\end{align}\]

Sum of \(n\) terms of AP series,

\[\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{51}} &= \frac{{51}}{2}\left[ {2 \times 10 + \left( {51 - 1} \right)4} \right]\\ &= \frac{{51}}{2}\left[ {20 + 50 \times 4} \right]\\ &= \frac{{51}}{2} \times 220\\ &= 51 \times 110\\ &= 5610\end{align}\]