Question

The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.

Answer 1

Consider the numbers are a , a+d , a+2d

Given that S₃ =12

⇒ a + a + d + a + 2d = 12

3(a + d) =12

a + d = 4

a = 4 - d ------------(1)

And sum of their cubes is 288.

(a)³ + (a + d)³ + (a + 2d)³ = 288

a³ + a³ +d³ +3a²d +3ad² +a³ + 8d³ + 6a²d +12ad² =288

3a³ + 9d³ +9a²d +15ad² =288

3(4-d)³ + 9d³ +9(4-d)²d +15(4-d)d² = 288 [using equation 1]

3(64 -d³ -48d +12d² ) + 9d³ + 9(16 + d² -8d) d + (60 -15d)d² = 288

192 - 3d³ - 144d + 36d² +9d³ + 144d + 9d³ - 72d² +60d² - 15d³ = 288

24d² = 288 -192

24d² =96

d² = 96/24

d² = 4

d = ±2

For d = 2, a = 4 – d = 4 – 2 = 2

The numbers will be 2, 4 and 6.

For d = - 2, a = 4 - (-2) = 4 + 2 = 6

The numbers will be 6, 4 and 2.

Hence, the required numbers are 2, 4 and 6.