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| 1. |
The sum of three numbers In A.P .is 12 and the sum of their cubes is 288. Find the numbers |
| Answer» Consider the numbers are a , a+d , a+2dGiven that S3\xa0=12⇒ a + a + d + a + 2d = 123(a + d) =12a + d = 4a = 4 - d ------------(1)And sum of their cubes is 288.(a)3\xa0+ (a + d)3\xa0+ (a + 2d)3\xa0= 288a3\xa0+ a3\xa0+d3\xa0+3a2d +3ad2\xa0+a3\xa0+ 8d3\xa0+ 6a2d +12ad2\xa0=2883a3\xa0+ 9d3\xa0+9a2d +15ad2\xa0=2883(4-d)3\xa0+ 9d3\xa0+9(4-d)2d +15(4-d)d2\xa0= 288 [using equation 1]3(64 -d3\xa0-48d +12d2\xa0) + 9d3\xa0+ 9(16 + d2\xa0-8d) d + (60 -15d)d2\xa0= 288192 - 3d3\xa0- 144d + 36d2\xa0+9d3\xa0+ 144d + 9d3\xa0- 72d2\xa0+60d2\xa0- 15d3\xa0= 28824d2\xa0= 288 -19224d2\xa0=96d2\xa0= 96/24d2\xa0= 4d\xa0= ±2\xa0For\xa0d\xa0= 2,\xa0a\xa0= 4 –\xa0d\xa0= 4 – 2 = 2The numbers will be 2, 4 and 6. For\xa0d\xa0= - 2,\xa0a\xa0= 4 - (-2) = 4 + 2 = 6The numbers will be 6, 4 and 2. Hence, the required numbers are 2, 4 and 6. | |