Question

sum of 3 numbers in ap is -3 and their product is 8 find the numbers​

Answer 1

Answer:

Step-by-step explanation:

Let a-d,a,a+d three numbers in AP

(a-d)+(a)+(a+d) = -3

3a = -3;

a = -3/3 = -1

Given that their product is 8

(a-d)(a)(a+d) = 8

($a^{2} -d^{2}$)(-1) = 8

($a^{2} -d^{2}$) = 8/-1 = -8

(1 - $d^{2}$) = -8

$d^{2}$ = -8-1; $d^{2}$ = -9

d = ±3

a = -1,d = 3     and a = -1,d = -3

a-d,a,a+d        and     a-d,a,a+d

-4,-1,2              and       2,-1,-4

Answer 2

Step-by-step explanation:

let the number be a-d a a+d and then solve