Question

Question 18
The sum of first five multiples of3 is :

A) 45
B) 55
C) 65
D) 75

Answer 1

The first five multiples of 3 are 3, 6, 9, 12 and 15.

Here, first term, a = 3,
Common difference, d = 3
Number of terms, n = 5
$\therefore S_5=\frac{5}{2}[2a+(5-1\left)d\right][\because S_n=\frac{n}{2}\{2a+(n-1)d\left\}\right]$
$=\frac{5}{2}[2\times 3+4\times 3]$
$=\frac{5}{2}(6+12)=5\times 9=45$