The sum of three numbers in A.P. is 12 and sum of their cubes is 288. Find the numbers.

#### Solution 1

Let the three numbers be *a *− *d*, *a* and* a* + *d*.

*S*_{3 }= 12

*a *− *d* + *a + a* + *d = *12

⇒ 3*a* = 12

⇒ *a* = 4 .....(1)

(a−d)^{3}+a^{3}+(a+d)^{3}=288

⇒(4−d)^{3}+4^{3}+(4+d)^{3}=288

⇒64+d^{3}+48d+12d^{2}+64+64−d^{3}−48d+12d^{2}=288

⇒24d^{2}=96

⇒d=±2

If *d* = 2, then the numbers are (4 − 2), 4 and (4 + 2), that is, 2, 4 and 6.

If *d* = −2, then the numbers are (4 + 2), 4 and (4 − 2), that is, 6, 4 and 2

#### Solution 2

In the given problem, the sum of three terms of an A.P is 12 and the sum of their cubes is 288.

We need to find the three terms.

Here,

Let the three terms b (a - d), a, (a + d) where a is the first term and d is the common difference of the A.P

So,

(a - d) + a + (a + d) = 12

3a = 12

`a= 12/3`

a = 4

Also, it is given that

`(a - d)^3 + a^3 + (a + d)^3 = 288`

So, using the properties:

`(a - b)^3 = a^3 - b^3 + 3ab^2 - 3a^2b`

`(a + b)^3 = a^3 + b^3 + 3ab^2 + 3a^2b`

We get

`(a - d)^3 + a^3 + (a + d)^3 = 288`

`a^3 - d^3 - 3a^3d^2a + a^3 + a^3 + d^3 + 3a^2d + 3d^2a = 288`

`3a^3 + 6d^2a = 288`

`a^3 + 2d^2 a = 96`

Further solving for *d *by substituting the value of *a*, we get,

`(4)^3 + 2d^2 (4) = 96`

`64 + 8d^2 = 96`

`8d^2 = 96 - 64``

`d^2 = 32/8`

On further simplification we get

d = 4

`d= sqrt4`

`d = +2`

Now, here d can have two values +2 and -2

So, on substituting the values of a = 4 and d = -2 in three terms we get

First term = a - d

So,

a - d = 4 - 2

= 2

Second term = a

So,

a = 4

Third term = a + d

So,

a + d = 4 + 2

= 6

Also on substituting the values of a = 4 and d = -2 in three terms, we get

First term = a - d

So,

a - d = 4 - (-2)

= 4 + 2

= 6

Second term = a

So,

a = 4

Third term = a + d

So,

a + d = 4 + (-2)

= 4 - 2

= 2

Therefore the three terms are 2, 4 and 6 or 6, 4 and 2