Question

Find three consecutive odd integers such that the sum of the first is two less than the second, and three more than the third is $70$.

Answer 1

Let the $3$ consecutive odd integer be $;x,x+2,x+4$

$x+2(x+2)+3(x+4)=70$

$x+2x+4+3x+12=70$

$6x+16=70$

$6x=54$

$x=9$

$\therefore 3$ no : are : $9,11,13$