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The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. find the numbers. |
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Answer» HI ,, LET ( a - d ), a , ( a + d ) are THREE terms in A .P According to the problem given , a - d + a + a + d = 12 3a = 12 a = 12/3 a = 4 ---( 1 ) ( a - d )³ + a³ + ( a + d )³ = 288 a³ -3a²d + 3ad² - d³ + a³ + a³ + 3a²d + 3ad² + d³ = 288 3a³ + 6ad² = 288 3 × 4³ + 6 × 4 d² = 288 [ from ( 1 ) ] 192 + 24d² = 288 24d² = 288 - 192 24d² = 96 d² = 96/24 d² = 4 d = ± 2 Therefore Required numbers are , 1 ) if a = 4 , d = 2 ( 4 - 2 ) , 4 , ( 4 + 2 ) 2 , 4 , 6 2 ) a = 4 , d = -2 Required numbers are , 6 , 4 , 2 I hope this helps you. : ) |
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