Question

Find the sum of integers divisible from 1 to 100 that are divisible by both 3 and 4 is

• A. 436
• B. 432
• C. 424
• D. 426

Answer 1

The correct option is B

432

The sequence which is divisible by 3 is 3,6,9,(12),15,18,21,(24),27,.....
And by 4 is 4,8,(12),16,20,(24),28.
First term = 12
The numbers are 12,24....96
$\Rightarrow$ 8 terms
S=$\frac{n}{2}(a+r)$

[1=96, a=12, n=8]

$\Rightarrow S=\frac{8}{2}[96+12]$
$=4x108$
$=432$