# Ex.5.3 Q4 Arithmetic Progressions Solution - NCERT Maths Class 10

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## Question

How many terms of the AP. $$9, 17, 25 \dots$$ must be taken to give a sum of $$636$$?

Video Solution
Arithmetic Progressions
Ex 5.3 | Question 4

## Text Solution

What is Known?

The AP and sum.

What is Unknown?

Number of terms.

Reasoning:

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

Where $$a$$ is the first term, $$d$$ is the common difference and $$n$$ is the number of terms.

Steps:

Given,

• First term, $$a = 9$$

• Common difference, $$d = 17 - 9 = 8$$

• Sum up to nth terms, $${S_n} = 636$$

We know that sum of $$n$$ terms of AP

\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\636 &= \frac{n}{2}\left[ {2 \!\times \!9 \!+\! \left( {n - 1} \right)8} \right]\\636 &= \frac{n}{2}\left[ {18 + 8n - 8} \right]\\636& = \frac{n}{2}\left[ {10 + 8n} \right]\\636& = n\left[ {5 + 4n} \right]\\636& = 5n + 4{n^2}\\4{n^2} + 5n - 636 &= 0\\\begin{bmatrix} 4{n^2} + 53n - \\48n - 636\end{bmatrix} &= 0\\\begin{bmatrix}n\left( {4n + 53} \right) -\\ 12\left( {4n + 53} \right)\end{bmatrix} &= 0\4n + 53)(n - 12) &= 0\end{align} Either \(4n + 53 = 0 or $$n - 12 = 0$$

$$n = - \frac{{53}}{4}$$ or $$n = 12$$

$$n$$ cannot be $$\frac{{ - 53}}{4}$$. As the number of terms can neither be negative nor fractional, therefore, $$n = 12$$

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