# The sum of three numbers in an A . P. is 27 and their product is 405. Find the numbers.

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### 2 Answers

The three numbers are in AP, so we can denote them as x , x + a and x + 2a

Now x + x + a + x + 2a = 27

=> 3x + 3a = 27

=> x + a = 9

Their product is equal to x ( x +a ) (x +2a ) = 405

substitute x = 9 - a

=> ( 9 - a) ( 9 ) ( 9 - a + 2a ) = 405

=> ( 9 - a ) ( 9) ( 9 + a) = 405

=> 81 - a^2 = 45

=> a^2 = 36

=> a = -6 or + 6

For a = -6, x = 15

For a = + 6 , x = 3

**The three numbers are 3 , 9 and 15.**

Let's note the numbers as:

a1,a2,a3.

The sum is:

a1 + a2 + a3 = 27

a1*a2*a3 = 405

Since the numbers are the consecutive terms of an a.p., we could write them:

a3 = a1 + 2d

a2 = a1 + d

d - the common difference

a1 + a1 + d + a1 + 2d = 27

We'll combine like terms:

3a1 + 3d = 27

We'll divide by 3:

a1 + d = 9 (1)

a1 = 9 - d

a1*a2*a3 = 405

a1(a1 + d)(a1 + d + d) = 405

a1*9*(9+d) = 405

We'll divide by 9:

a1(9+d) = 45

9a1 + a1*d = 45

9( 9 - d) + ( 9 - d)*d = 45

( 9 - d)(9 + d) = 45

81 - d^2 = 45

d^2 = 81 - 45

d^2 = 36

d = 6

a1 = 9 - 6

a1 = 3

a2 = 3 + 6

a2 = 9

a3 = 9 + 6

a3 = 15

For d = -6

a1 = 9 + 6

a1 = 15

a2 = 15 - 6

a2 = 9

a3 = 9 - 6

a3 = 3