Question

The sum of three consecutive terms is an AP is 6.and their product is -120. Find the three numbers please help

Answer 1

the three consecutive numbers are a,a+d,a+2d

a+a+d+a+2d=6

3a+3d=6

a+d=2

now,

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

a+d=2,(as we found in step 1)

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

(from first result a+d=2

d=2-a)

2a*(a-2(2-a)=120

2a*(4-a)=120

2a²-8a-120

thus

a=10

a=-6

consider only 10

a+d=2

d=2-a

d=-8

so the three no.s are 10,10-8,10-16

=10,2,-6

or

a=-6

d=8

so the numbers are

-6,2,10

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