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


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.  


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


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

From $(ii)$, we get




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


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

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

$\Rightarrow 8d^2=32$

$\Rightarrow d^2=4$

$\Rightarrow d=\pm 2$

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 $2$ . 


Simply Easy Learning

Updated on: 10-Oct-2022


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