Question

Sum of three consecutive numbers in ap is 9 and their product is 24

Answer 1

Let the A.P be a-d,a,a+d
a-d+a+a+d = 9
3a = 9
a=9/3
a=3

(a-d)×a×(a+d )=24
putting a=3
(3-d)×3×(3+d)=24
(3-d)(3+d)= 24/3
9-d²=8
-d² = 8-9
-d² = -1
d=1
Three consecutive numbers are
a-d = 3-1=2
a=3
a+d= 3+1 =4
Three consecutive numbers are 2 , 3, 4