# 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.

Video Solution
Arithmetic Progressions
Ex 5.3 | Question 8

## 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}

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