Question

find three numbers in arithmetic progression whose sum is 9 and the product is -165​

Answer 1

let the three numbers be a-b , a , a+b

according to first condition,

a-b+a+a+b=9

3a=9

a=3

according to second condition,

(a-b)(a)(a+b) = -165

a2-b2 x a = -165

a3-b2=-165

3 x 3 x 3 - b2 = -165

27-b2 = -165

b2 = 27+165

b2 = 192

b=

$\sqrt{192}$

substitute and get the number

a-b= 3-192 squre root

a = root 192

a+b = 3 + root 192

Answer 2

LET NOS BE

a-d, a, a+d

sum is 9

therefore,

a-d+a+a+d=9

3a=9. (because opposite sign get cancle)

a=9/3

a=3. ........equation 1

product is -165

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

(a^2-d^2) a =-165 (property (a+d) (a-d)=a^2-d^2)

a^3-ad^2= -165

putting value of 'a' from equation 1

(3)^3-3+×d^2 =-165

3 cube is 27

27-3d^2= -165

-3d^2= -165-27

-3d^2= -192

-d^2= -192/3

-d^2= -64

negative sign cancle

d^2 = 64

8 square is 64

therefore,

d= 8.............equation 2

THEREFORE NOS ARE

a-d

= 3-8

=-5

a= 3

a+d

=3+8

=11

therefore nos are

-5, 3, 11

HOPE IT HELPED MARK AS BRAINLIEST