If we consider a set of three numbers a, b and c which are in AP, we know that a + c =2b. Now, if they are in GP, a*c =b^2.

Now we have to see if it is possible to have three numbers satisfy both the equations.

If we express a as 2b - c and replace it in a*c = b^2, it gives us (2b-c)*c = b^2

=> 2bc – c^2 = b^2

=> b^2 + c^2 – 2bc =0

=> (b-c) ^2 =0

=> b = c

Also, as a = 2b-c = 2c-c = c

we find that all the three number have to be equal.

**So a set of three numbers which are in AP as well as GP are a set of three equal numbers.**

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.