Question

the sum of three numbers in ap is 12 and sum of the cubes is 288 find the numbers

Answer 1

Let the numbers be a-d, a and a+d

According to question

a - d + a + a + d = 12

=> 3a = 12

=> a = 4

Now,

(a-d)^3 + a^3 + (a+d) ^3 = 288

=> (4 - d) ^3 + 4^3 + (4+d)^3 = 288

=> 64 - d^3 - 12d ( 4-d) + 64 + 64 + d^3 + 12d(4+d) = 288

=> 64 + 64 + 64 + 48d + 12d^2 + 12d^2 - 48d = 288

=> 192 + 24d^2 = 288

=> 24d^2 = 96

=> d^2 = 4

=> d = 2


Required numbers are 2, 4, 6