Question

Hey

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

Answer 1

n = 40
a = 6
Last term = a40 = 40*6 = 240

this series is an A.P
6,12,.............240

we know that
sum = n/2 ( first term + last term)
sum = 40/2 ( 6 + 240)
sum = 20*( 246)
sum = 4920 ans

Answer 2

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