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

## Question

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

## 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, \(n = 12\)