Can a set of three numbers be in AP as well as GP?

Let us see algebraically, whether this is
possible.

Let a, b and c  be 3 numbers which are in both AP
and GP.

Since the numbers are in AP, a+c =
2b.........(1)

Since the numbers are in GP , ac = b^2
......(2).

From (1), c = 2b-a, Substitute  2b-c for c in
(2):

a(2b-a) = b^2.

Therefore
b^2- 2ab + a^2 = 0

Or  (b-a)^2 =
0.

Or a = b.

Or  (a,b,c ) =
(a,a,a,).

That implies the common diffrence = a-b = 0 and
the common ratio is b/a = 1.

Under the special case, a,a
and a are an AP with common diffrence d = 0 and a,a and a are in GP with common
diference d =1.