Question

pls help me to solve this sum..,

Answer 1

hi mate here is ur answer

let the terms of the AP be a , a+d ,a+2d

given to us is that their sum is 12 that means

a+a+d + a + 2d = 12

3a + 3d = 12

take 3 as common

we get,

3(a+d) = 12

a + d = 4

now since sum is 12 nos can be (1,4,7) (2,4,6)and (3,4,5)

but given to us is like sum of cubes is 408

so by cubing we know that

the nos. are (1,4,7)

hope it helps!!!

Answer 2

Hey friend, Harish here.

Here is your answer.

Let the numbers in A.P be : (a - d) , a , (a + d)

Here: d - is the common difference.  

Given that sum of the three numbers = 12 .

⇒ (a - d) + a + (a + d) = 12

⇒ 3a = 12

⇒ a = 4

Also it is given that,

Sum of Cube of the numbers = 408

⇒ (a - d)³ + (a)³ + (a + d)³ = 408

⇒ (4 - d)³ + (4)³ + (4 + d)³ = 408

⇒ [ 64 - d³ - 48d + 12d² ] + 64 + [ 64 + d³ + 48d + 12d² ] = 408

⇒ 24d² + 192 = 408

⇒ 24d² = 216

⇒ d² = 9

⇒ d = 3

So,  (a - d) = 4 - 3 = 1  ;  a = 4 ; (a + d) = 4 +3 = 7

∴ The three numbers in AP are: 1 , 4 , 7

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Hope my answer is helpful to you.