Question

The sum of three numbers in AP is 27 and their product is 405. Find the numbers.

Answer 1

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) => (a^2 - d^2)(a) = 405

[(9)^2 - d^2] (9) = 405

(81 - d^2) (9) = 405

729 - 9d^2 = 405

9d^2 = 729 - 405

9d^2 = 324

d^2 = 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.

VERIFICATION :

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

3 + 9 + 15 = 27

27 = 27

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

(3)(9)(15) = 405

405 = 405

Hence verified.........

Answer 2

the numbers in AP are a-d,a,a+d.

a-d+a+a+d=27

=> 3a = 27

=> a = 9

since product of three numbers is 405, we have,

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

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

=> (81-$d^{2}$) 9 = 405

=> 81-$d^{2}$ = 405/9

=> 81-$d^{2}$= 45

=> $d^{2}$ = 81-45

=> $d^{2}$= 36

=> d = ₊⁻ 6

thus, the numbers are

3,9,15

or

15,9,3