1.

The sum of first 40 positive integers divisible by 6 is (a) 2460 (b) 3640 (c) 4920 (d) 4860

Answer»

Correct answer is (c) 4920

The positive integers divisible by 6 are 6, 12, 18,....

This is an AP with a = 6 and d = 6.

Also, n = 40 (Given)

Using the formula, \(S_n=\frac{n}{2}[2a+(n-1)d],\) we get

\(S_{40}=\frac{40}{2}[2\times6+(40-1)\times6]\)

= 20(12 + 234)

= 20 x 246

= 4920

Thus, the required sum is 4920



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