Question

Sum of three numbers in a.p. is 12 and sum of their cubes is 288. Find the numbers.
Please solve this

Answer 1

Answer.


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 =2883(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.

Answer 2

Refer to the image below for the solution