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


Given:

Three numbers are in A.P. 

The sum of these numbers is 12 and the sum of their cubes is 288.

To do:

We have to find the numbers.  

Solution:

Let the first three terms of the AP be $a−d,\ a,\ a+d$.

According to the problem,

$a−d+a+a+d=12\ ......( i)$

$( a−d)^3+(a)^3+( a+d)^3=288\ .....(ii)$

From $(i)$, we get

$3a=12$

$\Rightarrow a=\frac{12}{3}=4$

From $(ii)$, we get

$a^3-d^3-3ad(a-d)+a^3+a^3+d^3+3ad(a+d)=288$

$3a^3-3a^2d+3ad^2+3a^2d+3ad^2=288$

$3a^3+6ad^2=288$.....(iii)

On putting $a=4$ in equation $(iii)$, we get,

$3(4^3+2(4)d^2)=3\times96$

$\Rightarrow 64+8d^2=96$

$\Rightarrow 8d^2=96-64$

$\Rightarrow 8d^2=32$

$\Rightarrow d^2=4$

$\Rightarrow d=\pm 2$

This implies,

If $a=4, d=2$ then

$a-d=4-2=2, a=4, a+d=4+2=6, a+2d=4+2(2)=4+4=8$

The required A.P is $2, 4, 6, 8,......$

If $a=4, d=-2$ then

$a-d=4-(-2)=4+2=6, a=4, a+d=4+(-2)=2, a+2d=4+2(-2)=4-4=0$

The required A.P is $6, 4, 2, 0,......$

The numbers are $2, 4$ and $6$ or $6, 4$ and $2$ . 

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Updated on: 10-Oct-2022

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