Question

Sum of three consecutive terms in AP is 6 and their product is
120 . find three terms.​

Answer 1

Answer:

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

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Answer 2

Answer:

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

3a = 6

A = 2

(a-d)×a×(a+d) = 120

(2-d×)2×(2+d) = 120

(4 - d^2) 2 = 120

4 - d^2 = 60

d^2 = - 64

D = 8

AP = - 6,2,10