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The sum of three numbers in A.P. is 12 and the sum of their cubes is 408, then the 1st three numbers of A.P. are 1, 4 ,7.

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Detailed Solution

Given that,
The sum of three numbers in A.P. is 12 and the sum of their cubes is 408.
Let the three numbers which are in A.P. as a-d, a, a+d
Therefore,

a − d + a + a + d = 12 ⇒ 3 a = 12 ⇒ a = 123 ⇒ a = 4

Now,

{(a - d)}^{3} + a^{3} + {(a + d)}^{3} = 408

\Rightarrow a^{3} - 3 a^{2} d + 3 a d^{2} - d^{3} + a^{3} + a^{3} + 3 a^{2} d + 3 a d^{2} + d^{3} = 408

\Rightarrow 3 a^{3} + 6 {ad}^{2} = 408

Put the value of a,

3 × 43 + 6 ⋅ 4 ⋅ d 2 = 408 ⇒ 3 × 64 + 24 d 2 = 408 ⇒ 192 + 24 d 2 = 408 ⇒ 24 d 2 = 408 − 192

⇒ 24 d 2 = 216 ⇒ d 2 = 21624 ⇒ d 2 = 9 ⇒ d = 3

Therefore, the three numbers are,
first term is:

a − d = 4 − 3 ⇒ a − d = 1

second term is a=4 and
the third term is:

a + d = 4 + 3 ⇒ a + d = 7

The three numbers which are in A.P. are 1, 4, and 7.
Hence, the statement is true.
Hence, option 1 is correct.

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