Question:
Choose the correct answer in each of the following questions:
The sum of first 40 positive integers divisible by 6 is
(a) 2460
(b) 3640
(c) 4920
(d) 4860
Solution:
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, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we get
$S_{40}=\frac{40}{2}[2 \times 6+(40-1) \times 6]$
$=20(12+234)$
$=20 \times 246$
$=4920$
Thus, the required sum is 4920.
Hence, the correct answer is option C.