Question

find four numbers in AP whose sum is -2 and product is 40

Answer 1

now,
you can proceed by using formula
(a-3d)(a-d)(a+d)(a+3d)
put value of a and find d

Answer 2

Let the numbers in the Ap be a - d,a + d,a - 3d,a + 3d.

Given that the sum is -2.

a - d + a + d + a - 3d + a + 3d = -2

4a = - 2

a = -2/4

a = -1/2.


Given that product is 40.

(a-d)(a+d)(a-3d)(a+3d) = 40

We know that (a+b)(a-b) = a^2 - b^2.

(a^2 - d^2)(a^2 - 9d^2) = 40

(a^2*a^2 - a^2*9d^2 - d^2a^2 + d^2*9d^2) = 40

a^4 - 10a^2d^2 + 9d^2d^2 = 40

a^4 - 10a^2d^2 + 9d^4 = 40

(-1/2)^4 - 10 * (-1/2)^2 * d^2 + 9d^4 = 40

1/16 - 5/2 * d^2 + 9d^4 = 40

9d^4 - 5/2d^2 + 1/16 = 40

9d^4 - 5/2d^2 + 1/16 - 40 = 0

9d^4 - 5/2d^2 - 639/16 = 0

LCM = 16.

144d^4 - 40d^2 - 639 = 0

144d^4 - 324d^2 + 284d^2 - 639 = 0

36d^2(4d^2 - 9) + 71(4d^2 - 9) = 0

(36d^2 + 71)(4d^2 - 9) = 0 

(36d^2 + 71)(2d + 3)(2d - 3) = 0

(36d^2 + 71) = 0, 2d + 3 = 0, 2d - 3 = 0
    
                            d = -3/2, d = 3/2.


Forget About (36d^2 + 71) = 0, It will be d = iroot71/6, -i root 71/6.

d = 3/2, -3/2

Since d cannot be negative, So d = 3/2.

Therefore the four numbers are:

a - d = -1/2 - 3/2

        = -4/2

        = -2.


a + d = -1/2 + 3/2

         = -1 + 3/2

         = 2/2

         = 1.



a + 3d = -1/2 + 3(3/2)

           = -1 + 9/2

           = 8/2

           = 4.



a - 3d = -1/2 - 3(3/2)

           = -1/2 - 9/2

           = -10/2

           = -5.



Therefore the four numbers in the AP are -5,-2,1,4.


Hope this helps!