Question

the sum of three numbers in AP is 12 and the sum of their cubes is 288. find those numbers.

Answer 1

the no. are in ap so it can be a+(n-1)d a+(2n-1)d , a+(3n-1)d

therefore,  according to the question =  a+2a+3a =12

a=2  

hence the no. are 2 , 4 , 6 and we can verify,

(2)³ + (4)³ + (6)³=288

Answer 2

Let the first term be a-d

Let the common difference be d

Let the AP be a-d,a,a+d

Their sum is 12

a-d+a+a+d=12

==) 3a=12

==) a=12/3

==) a=4

Their sum of cubes is 288

(a-d)³+a³+(a+d)³=288

(4-d)³+4³+(4+d)³=288..............(1)

(a-b)³+(a+b)³

=a³+b³+3a²b+3ab²+a³-b³-3ab²+3ab²

=2a³+6ab²

Here (4-d)³+(4+d)³=2*4³+6*4*d²

=128+24d²+64=288

=192+24d²=288

=24d²=288-192

=24d²=96

=d²=96/24

=d=2 or -2

We found a to be 4.

The APs can be 4-2,4,4+2

=2,4,6

or when d=-2

Ap is 4+2,4,4-2.

=6,4,2

Hope it helps you.