Question

The sum of three consecutive terms in AP is 6and their product is -120

Answer 1

Let the 3 terms of AP is (a+d),(a),(a-d)

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

3a=6

a=2 ......(1)

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

(a²-d²)*a = -120

Sub. a from 1

[(2)²-d²]*2=-120

4-d²=-120/2

4-d²= -60

-d²= -60-4

d²=64

d = 8

Answer 2

Let the terms be

a-d, a, a+d

Where a is your first term

d is the common difference

a-d+a+a+d=6

3a=6

a=2 --- (1)

a(a+d)(a-d)= -120

a^3 -ad^2= -120

From (1)

a=2

8-2d^2= -120

d^2= (-120-8)÷-2

d^2= 64

d= 8

So, the 3 terms are:

1) a-d = -6

2) a = 2

3) a+d = 10