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

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## 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?

Video Solution
Arithmetic Progressions
Ex 5.3 | Question 6

## 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 $$6973.$$

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