Question

Find the sum of the first forty positive integers divisible by 6.

Answer 1

Hint: Here they have asked us to find the sum of those numbers which are divisible by 6 and in a continuous manner. These divisible numbers are nothing but the multiples of 6 with a common difference between consecutive terms. So we can use the formula for A.P. for finding the sum of these numbers.

Complete step by step answer:

We have to find the sum of the first forty positive integers divisible by 6.
The very first number divisible by 6 is that number itself.
Now next to it are nothing but its multiples only. That is 12, 18, 24….
But we need to find only the first forty numbers.
If we observe there is a common difference between the integers divisible by 6.
Thus it forms an A.P.
In this A.P.,
First term a =6
Common difference d= 6
Number of terms n=40
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
\[
   \Rightarrow \dfrac{{40}}{2}\left[ {2 \times 6 + \left( {40 - 1} \right)6} \right] \\
   \Rightarrow 20\left[ {12 + 39 \times 6} \right] \\
   \Rightarrow 20\left[ {12 + 234} \right] \\
   \Rightarrow 20 \times 246 \\
   \Rightarrow 4920 \\
\]
Therefore the sum of the first forty positive integers divisible by 6 is 4920.

Note: Generally students find the first forty numbers that are divisible by 6 and add them. But it consumes time. So if you find that the series is having anything special like it is an A.P., G.P., or H.P. then you can directly use the formula to find the sum.
Additional information: An arithmetic progression is a series of numbers in which the numbers are having a common difference d between them. If a is the first term then the series a+d,a+2d,….forms an A.P.