Question

Find four numbers in AP such that their sum is 24 and product is 945.​

Answer 1

Answer:

Let the terms be (a - 3d), (a - d), (a + d) and (a + 3d)

Given : Sum of the numbers in AP = 24

So, 

a - 3d + a - d + a + d + a + 3d = 24

4a = 24

a = 6

And the product of these four numbers is 945

So,

(a - 3d) (a - d) (a + d) (a + 3d) = 945

Putting the value of a =6

(6 - 3d) (6 - d) (6 + d) (6 + 3d) = 945

(36 - 9d²) (36 - d²) = 945

1296 - 360d² + 9d⁴ = 945

9d⁴ - 360d² + 1296 - 945 = 0

9d⁴ - 360d² + 351 = 0   Dividing it by 9 we get

d⁴ - 40d² + 39 = 0

d⁴ - 39d² - d² + 39 = 0

d²(d² - 39) - 1(d² - 39) = 0

(d² - 1) (d² - 39) = 0

d² - 1 = 0

d² = 1

d = √1

d = 1

d² - 39 = 0

d² = 39

d = √39

d = 6.244  

d = 6.244 is not possible, so d = 1

So, a = 6 and d = 1

The required four numbers of the AP are (6 - 3), (6 - 1), (6 + 1) and (6 + 3)

= 3, 5, 7 and 9

Answer 2

Answer:

let no be a-3d, a-d, a+d,a+3d

a-3d+a-d+a+d+a+3d=24

4a=24

a=6

SUBSTITUTING VALUE OF a:

6-3d*6-d*6+d*6+3d=945

(36-d^2)*(36-9d^2)=945

1296-324d^2-36d^2+9d^4=945

LET d^2be a

9a^2-360a+351=0. (DIVIDING BY 9)

a^2-40a+39=0

a^2-39a-a+39=0

a(a-39)-1(a-39)=0

a=39,1

SUBSTITUTING a=d^2

d=1,+6√3,-6√3

if d=1 then :

3,5,7,9

if d=6√3 then :

6(1-3√3),6(1-√3),6(1+√3),6(1+3√3)

if d=-6√3 then:

6(1+3√3),6(1+√3),6(1-√3),6-3√3)