Question

The sum of the three consecutive number in A.P. is 21 and product is 231? find the nubers

Answer 1

Let the numbers be a-d, a, a+d

According to question
a-d+a+a+d = 21
=> 3a = 21
a = 7

Now,

Product = 231
(a-d) (a) (a+d) = 231
=> (7-d)(7+d) * 7 = 231
=> 49 - d^2 = 33
=> d^2 = 49-33
=> d^2 = 16
d = 4


Required numbers are 3, 7, 11

Answer 2

let them be (x-d),x,(x+d)
x=21/3=7
(7-d)*7*(7+d)=231
(49-d²)*7 =231
343-7d² =231
7d²=343-231
7d²=112
d²=16
d=4

answer the numbers are
3,7,11