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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Answered by
37
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.
: )
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.
: )
ShaikElijah:
thank u
Answered by
1
Answer:
Step-by-step explanation:
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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