Question

Find three numbers in arithmetic progression whose sum is 9 and product is -165

Answer 1

let the AP be a-d,a,a+d

sum of AP = a-d+a+a+d=9

=> 3a = 9

=> a = 3

product of AP:-

(3-d)(3)(3+d) = -165

9-d² = -55

=> d²= 64

so d = 8

so AP is -5,3,11...

I hope it helps

Answer 2

let 3 numbers be (a-d)(a)(a+d)

A.T.Q

a-d+a+a+d=9

3a=9

a=3

now,

(a-d)(a)(a+d)=-165

(a-d)(a+d)×3=-165

a²-d²=-165/3

3²-d²=-55

9-d²=-55

9+55=d²

64=d²

d=±√64

d=±8

so,

the 3 numbers with a=3 & d=8

3,11,19

OR

the 3 numbers with a=3 & d=-8

3,-5,-13

hope this answer will help you.