Question

Divide 39 in 3 parts in a.p. such that the largest exceeds the smallest by 6.

Answer 1

Let the smallest term be a and the common difference be d.
Now,
 (a+2d)-a=6
 2d=6
 d=3

Note that the sum has been given as 39.
We need to divide 39 into 3 parts in ap, we also have the common difference being 3, therefore,
 a+a+d+a+2d=39
 3a+3d=39
 3a+9=39
 3a=30
 a=10

Hence the ap would be,
 10, 13, 16