# Ex.5.3 Q12 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

Find the sum of first \(40\) positive integers divisible by \(6.\)

## Text Solution

**What is Known?**

Positive integers divisible by \(6\)

**What is Unknown?**

Sum of first \(40\) positive integers divisible by \(6,\) \({S_{40}}\)

**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:**

The positive integers that are divisible by \(6\) are \(6, 12, 18, 24, \dots\)

It can be observed that these are making an AP

Hence,

- First term, \(a = 6\)
- Common difference, \(d = 6\)
- Number of terms, \(n = 40\)

As we know that Sum of \(n\) terms,

\[\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{40}} &= \frac{{40}}{2}\left[ {2 \times 6 + \left( {40 - 1} \right)6} \right]\\ &= 20\left[ {12 + 39 \times 6} \right]\\ &= 20\left[ {12 + 234} \right]\\& = 20 \times 246\\ &= 4920\end{align}\]