Question

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

Answer 1

Solution :

The first positive integers divisible by 6 are 6, 12, 18, .... Clearly, it is an AP with first term a = 6 and common difference d = 6. We want to find $S_{40}$.

∴ $S_{40}$ = $\frac{40}{2}$ [2 × 6 + (40 - 1) 6]

                             = 20(12 + 39 × 6)

                             = 20(12 + 234) = 20 × 246 = 4920

Answer 2

Answer:

4920

Step-by-step explanation:

The first 40 positive integers divisible by 6 are 6, 12, 18, 24, 30, ......., 240.

Now, we have to find the value of the sum of the first 40 integers divisible by 6.

Hence, we have to get

6+ 12+ 18+ 24+ 30+ ......... +240

Taking 6 as common from all the terms, we get,

6(1+ 2+ 3+ 4+ ......+ 40)

Now, the sum within the brackets is the sum of the first 40 natural numbers.

Hence, 6+ 12+ 18+ 24 +30 + ....... +240

= 6(1+ 2+ 3+ 4+ ......+ 40)

= 6 [40(40+1)/2] {Sum of first n natural numbers is given by n(n+1)/2}

=4920 (Answer)