The sum of three numbers in AP is 12 and the sum of their cubes is 288. Find the numbers.
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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.
⇒ 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.
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