The sum of three numbers in A.P. is 12 and the

Solution:

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 be where, a is the first term and d is the common difference of the A.P

So,

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

$3 a=12$

$a=\frac{12}{3}$

$a=4$

Also, it is given that

$(a-d)^{3}+a^{3}+(a+d)^{3}=288$

So, using the properties:

$(a-b)^{3}=a^{3}-b^{3}+3 a b^{2}-3 a^{2} b$

 

$(a+b)^{3}=a^{3}+b^{3}+3 a b^{2}+3 a^{2} b$

We get,

$(a-d)^{3}+a^{3}+(a+d)^{3}=288$

$a^{3}-d^{3}-3 a^{2} d+3 d^{2} a+a^{3}+a^{3}+d^{3}+3 a^{2} d+3 d^{2} a=288$

$3 a^{3}+6 d^{2} a=288$

$a^{3}+2 d^{2} a=96$

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

$(4)^{3}+2 d^{2}(4)=96$

$64+8 d^{2}=96$

$8 d^{2}=96-64$

$d^{2}=\frac{32}{8}$

On further simplification, we get,

$d=4$

$d=\sqrt{4}$

$d=\pm 2$

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 = 

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  in three terms, we get,

First term = 

So,

$a-d=4-(-2)$

$=4+2$

 

$=6$

Second term = a

So,

$a=4$

Third term $=a+d$

$a=4$

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 .