Question

the sum of three consecutive terms in an ap is 6 and their product is - 120 find the three numbers

Answer 1

let the numbers be x, x+1,x+2
sum = -120
x+x+1+x+2= -120
3x+3=-120
3x=-120-3
3x=-123
x=-123/3
x= - 41
therefore
1 no =x= -41
2 no = x+1= -41+1=-40
3 no =x+2=-41+2=-39

Answer 2

solutions:-

Let the three terms be
 a-d,a,a+d  
sum=a-d+a+a+d=6
     =3a=6
     a=2

product =(a-d)a(a+d)=-120
            (a^2-d^2)a=-120
            (4-d^2)2=-120
             8-2d^2=-120
             -2d^2=-128
            d^2=64
                d=8

ap are -6,2 10