Question

The sum of 3 terms of ap is 60 and their product is 6000

Answer 1

let the 3 terms be (a-d),a,(a+d)

A/q

a-d+a+a+d=60

3a=60

a=20

now,

(a-d)a(a+d)=6000

(a^2-d^2)a=6000

(20^2-d^2)20=6000

400-d^2=300

400-300=d^2

100=d^2

so,d=-10 or +10

now ,a=20 and d=-10 or +10

hence 3 terms = (20-10),20,(20+10)

10,20,30

or

[20-(-10)],20,[20+(-10)]

(20+10),20,(20-10)

30,20,10