Question

The sum of there no. in AP is 27 and their product is 405 find the number

Answer 1

Answer:

Let the numbers be a - d, a, a + d which are in AP.

(a - d) + (a) + (a + d) = 27 .........(1)

(a - d)(a)(a + d) = 405 ..........(2)

(1) => 3a = 27

a = 9

(2) => (a2 - d2)(a) = 405

[(9)2 - d2] (9) = 405

(81 - d2) (9) = 405

729 - 9d2 = 405

9d2 = 729 - 405

9d2 = 324

d2 = 36

d = 6

Now, substitute the values in the 1st equation.

(a-d) = 9 - 6 = 3

a = 9

(a+d) = 9 + 6 = 15

Hence the numbers are 3, 9 and 15.