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

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## Question

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

Video Solution
Arithmetic Progressions
Ex 5.3 | Question 7

## 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}