Question

In a set of three consecutive natural numbers, the sum of the last two numbers is equal to three times the first number. Find the sum of all the three numbers.

Answer 1

Let the $3$ consecutive numbers be $a-d,a,a+d$

A/Q $\Rightarrow 3(a-d)=a+a+ad3a-3d=2a+d\Rightarrow a=4d$

$\therefore$ SUm of numbers is $a-d+a+a+d=3a=3\times 4d=12d$