Question

Find three consecutive numbers such that seven times the smallest number may be equal to three the sum of the other two.

Answer 1

Let the numbers be a, $a+1,a+2$.

$\Rightarrow 7\left(a\right)=3(a+1+a+2)$

$\Rightarrow 7a=3(2a+3)$

$\Rightarrow 7a=6a+9$

$\Rightarrow a=9$

Numbers $=9,10,11$.