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:

The positive integers that are divisible by 6 are

6, 12, 18, 24 …

It can be observed that these are making an A.P. whose first term is 6

and common difference is 6.

a = 6 and d = 6

sn=n/2(2a+(n-I) d)

S40 =?

= 20[12 + (39) (6)]

= 20(12 + 234)

= 20 × 246

= 4920

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