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

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## 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.

Video Solution
Arithmetic Progressions
Ex 5.3 | Question 5

## 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}

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