Question

The sum of three consecutive terms in an A.P is 6 and their product is -120. Find the three numbers.
'al -10,2,6
b) -6,2,10
c) -2,-6,-10 d) —2,6,10​

Answer 1

The three consecutive no. =-6, 2,10

Answer 2

Step-by-step explanation:

let the consecutive terms be

a-d,a,a+d

according to question

a+d+a+a-d = 6

3a = 6

therefore a = 2

situation 2

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

putting a = 3 in equation 2

(2+d)×2×(2-d) = -120

2(4 -d^2) = -120

8 - 2d^2 = -120

- 2d^2 = -128

d^2 = 64

therefore d = 8

therefore

number 1 = 10

number 2 = 2

number 3 = -6