Question

find the 3 terms in AP whose sum is 6 and whose sum of cube is 216

Answer 1

Let the 3 terms in an A.P. be a-d,a,a+d respectively.
As per 1st given condition,
a-d+a+a+d=6
3a=6
a=6/3
a=2.......................... (1)


As per 2nd given condition,
(a-d)3 + (a)3 + (a+d)3 = 216
(2-d)3 +(2)3 + (2+d)3 =216
now using the formulae of (a+b)3 &(a-b)3
(2)3-3* (2)2*d+3*2*d2-d3+8+(2)3+3*(2)2
*d+3*2*d2+d3=216

...............[(a+b)3=(a)3+3*a2*b+3*a*b2+b3 ] &
................[(a-b)3=(a)3-3*a2*b+3*a*b2-b3]

24+12d2=216
12d2=216-24
d2=192/12
d2=16
d=4

Now,
(a-d)=(2-4)
=-2
(a)=2
(a+d)=(2+4)
=6
Therefore, the 3 terms in an A.P. are
-2,2,6