The sum of three numbers in ap is 27 and thir products is 405 . Find the numbers
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Solution :
Let ( a - d ), a , ( a + d ) are three
numbers in A.P
i ) sum = 27
=> a - d + a + a + d = 27
=> 3a = 27
=> a = 27/3
=> a = 9 ---( 1 )
ii ) product = 405
=> ( a - d ) × a × ( a + d ) = 405
=> ( a² - d² ) a = 405
=> ( 9² - d² ) × 9 = 405
=> 81 - d² = 405/9
=> 81 - d² = 45
=> -d² = 45 - 81
=> - d² = - 36
=> d² = 6²
=> d = ± 6
Therefore ,
1 ) if a = 9 , d = 6
Required 3 terms are
a - d = 9 - 6 = 3
a = 9
a + d = 9 + 6 = 15
2 ) if a = 9 , d = -6
required 3 terms are
15 , 9 , 3
••••
Let ( a - d ), a , ( a + d ) are three
numbers in A.P
i ) sum = 27
=> a - d + a + a + d = 27
=> 3a = 27
=> a = 27/3
=> a = 9 ---( 1 )
ii ) product = 405
=> ( a - d ) × a × ( a + d ) = 405
=> ( a² - d² ) a = 405
=> ( 9² - d² ) × 9 = 405
=> 81 - d² = 405/9
=> 81 - d² = 45
=> -d² = 45 - 81
=> - d² = - 36
=> d² = 6²
=> d = ± 6
Therefore ,
1 ) if a = 9 , d = 6
Required 3 terms are
a - d = 9 - 6 = 3
a = 9
a + d = 9 + 6 = 15
2 ) if a = 9 , d = -6
required 3 terms are
15 , 9 , 3
••••
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