Question

the sum of three consecutive terms in an ap is six and their product is- 120 find the three numbers​

Answer 1

Let the 3 terms of tje AP be a-d, a and a+d where a is the first term and d is the common difference.

So (a-d)+(a)+(a+d) =6

3a = 6

a=2

Also, (a-d)*(a)*(a+d)= -120

(2-d)*2*(2+d) = -120

4 - d^2 = -60

d^2 = 64

d = 8 or -8

Hence the three numbers are:

-6, 2, 10