Question

find the sum of the first term 40 positive integer divisible by 6

Answer 1

First term= 6
Last Term= 240 [6×40]

Sum=40[6+240]÷2
Sum=20×246
Sum=4920

The sum of the first term 40 positive integer divisible by 6 is 4920

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