The sum of three numbers in A. P is 12 and the sum of their cubes is 408. Find the number.
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The numbers are 1,4,7
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Let three terms (a - d) , a , (a + d ) are in AP.
A/C to question,
sum of three terms = 12
(a - d ) + a + (a + d ) = 12
3a = 12 ⇒ a = 4
again, question said ,
sum of their cubes = 408
(a - d)³ + a³ + (a + d)³ = 408
a³ - 3a²d + 3ad² - d³ + a³ + a³ + 3a²d + 3ad² + d³ = 408
3a³ + 6ad² = 408
now put the value of a = 4
3(4)³ + 6(4)d² = 408
3 × 64 + 24d² = 408
24d² = 408 - 192 = 216
d² = 9 ⇒ d = ±3
Hence, terms are (a - d) , a , (a + d) :
when d = +3
(a - d ) = (4 -3) = 1
a = 4
(a + d ) = 3 + 4 = 7
e.g., three terms are : 1, 4 ,7
when we take d = -3
(a - d) = 4 - (-3) = 7
a = 4
(a + d) = 4 + (-3) = 1
e.g., three terms are : 7, 4 , 1
A/C to question,
sum of three terms = 12
(a - d ) + a + (a + d ) = 12
3a = 12 ⇒ a = 4
again, question said ,
sum of their cubes = 408
(a - d)³ + a³ + (a + d)³ = 408
a³ - 3a²d + 3ad² - d³ + a³ + a³ + 3a²d + 3ad² + d³ = 408
3a³ + 6ad² = 408
now put the value of a = 4
3(4)³ + 6(4)d² = 408
3 × 64 + 24d² = 408
24d² = 408 - 192 = 216
d² = 9 ⇒ d = ±3
Hence, terms are (a - d) , a , (a + d) :
when d = +3
(a - d ) = (4 -3) = 1
a = 4
(a + d ) = 3 + 4 = 7
e.g., three terms are : 1, 4 ,7
when we take d = -3
(a - d) = 4 - (-3) = 7
a = 4
(a + d) = 4 + (-3) = 1
e.g., three terms are : 7, 4 , 1
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