Question

The sum of three consecutive term in an A.P. is 6 and there product is-120 find the threel numbers

Answer 1

Answer:

let the three terms = a-d,a,a+d

ATQ

a-d+a+d+a=6

3a=6

a=2

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

put value of a

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

$4 - {d}^{2} = - 60 \\ { - d}^{2} = - 64 \\ d = + 8or - 8$

if d =8

so ap will be

-6,2,10

if d=-8

so ap will be

10,2,-6