Question

If sum of three numbers in.an A.P. is 12. And sum of their squares is 288. Then find the numbers.

Answer 1

let the nos be 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

from (1)

3(4-d)3 + 9d3 +9(4-d)2d +15(4-d)d2 = 288               

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.