Question

the sum of the three consecutive number in A.P is 21 and their products is 231. find the number

Answer 1

let the A.P be
a-d,a,a+d
addition will be ;
a-d + a + a+ d = 21
3a = 21
a= 7
a-d ×a × a +d = 231
(7-d ) × (7+d) ×7 =231
49 - d^3 = 33
d = 16

Now the no. will be
a-d = 7-16= -9
a. = 7
a+d = 16+7= 23
the no. are -9,7,23

Answer 2

Here is your solution


let,

Three numbers be a,a-d,a+d

so a+a-d+a+d = 21

a = 7

A/q
(a)(a-d)(a+d) = 231

(7)(7-d)(7+d) = 231

49- d2 = 231/7

d2 = 16

therefore d = +4 or - 4

so no(s) r : 3, 7 and 10 or 10, 7 and 3

hope it helps you