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

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### 2 Answers

There are 3 numbers a, b and c in AP.

Their sum a + b + c = 27

and their product a*b*c = 405

Also b = a + d and c = a + 2d

=> a + a + d + a + 2d = 27

=> 3a + 3d = 27

=> a + d = 9

=> b = 9

a*b*c = (b - d)*b(b + d) = 405

=> (b^2 - d^2)*9 = 405

=> (81 - d^2) = 45

=> d^2 = 36

=> d = 6

a = 9 - 6 = 3

c = 9 + 6 = 15

**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