If A,B,C are the terms of an AP then the middle terms is the arithmetical mean of the neighbor terms:
B = (A+C)/2
We'll cross multiply and we'll get:
2B = (A+B) (1)
If 2^A,2^B,2^C are the terms of a GP then the middle terms is the geometric mean of the neighbor terms:
2^B = sqrt(2^A*2^C)
We'll square raise both sides:
2^2B = 2^A*2^C
Since the bases of the exponentials from the right side are matching, we'll add the exponents:
2^2B = 2^(A+C)
Since the bases are matching, we'll apply one to one property:
2B = (A+B)
According to (1), the relation is true for any consecutive terms of an AP.
We'll choose the terms:
A = 2
We'll choose the common difference as d = 10
B = A + d
B = 2 + 10
B = 12
C = B + d
C = 12 + 10
C = 22
Now, as A, B and C are terms of an AP we know that A+C = 2B.
It is also given that 2^A, 2^B and 2^C are in GP. So we have 2^A*2^C = (2^B) ^2
=> 2^ (A+C) = 2^2B
Now for 2^ (A+C) to be equal to 2^2B, A+C = 2B. This is true for all values of A, B and C as they are in AP
As an example if we take A = 2, B = 5 and C = 8 which are in an AP with first term 2 and common difference 3, 2^A, 2^B and 2^C are equal to 4, 32, 256 respectively which are in a GP with the first term 4 and common ratio 8.
Any three values in AP will make 2^A, 2^B and 2^C form a GP.
Since A, B and C are in AP, the succesuve terms have the same common difference. So, B-A = C-B.
Therefore 2B = A+C.
Since 2^A, 2^B and 2^C are in GP, the successive tems have the same common ratio. Therefore 2^B/2^A = 2^C / 2^A .
(2^B)^2 = 2^A * 2^c
2^2B = 2^(A+C).
Bases being same, we equate the powers.
2^2B = A+C.
Similarly for x^A , x^B, and x^C ,
(x^B)^2 = x^A*x^C
x^(2B) = x^(A+C) holds as 2B = A+C.
Therefore for any integers in A,B and C in A P , we can choose a base x. Then x^A, x^B and x^C are in GP.
We take x = 7. A= 3, B = 5 and C = 7.
7^3 , 7^5 and 7^7 are in GP with 7^2 as common ratio.