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

Go back to  'Ex.5.3'

## Question

Find the sum of first $$40$$ positive integers divisible by $$6.$$

Video Solution
Arithmetic Progressions
Ex 5.3 | Question 12

## 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}

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school