How do we prove that 5^a, 5^b, 5^c, ... is a geometric sequence if it is known that a, b, c, ... is an arithmetic sequence?An arithmetic sequence is one for which the difference between consecutive terms is a constant, so b-a, c-b, ... all have the same value (call it r).
A geometric sequence is one for which the ratio between consecutive terms is a constant. So, to show that 5^a, 5^b, 5^c, ... is a geometric sequence, just look at the ratios: 5^b/5^a = 5^(b-a), 5^c/5^b = 5^(c-b), .... These ratios are all equal to the same value, namely 5^r (because b-a = c-b = ... = r). Therefore, 5^a, 5^b, 5^c, ... is a geometric sequence (whose common ratio is 5^r where r is the common difference of the arithmetic sequence).
Here's another way to think of this. In general, the nth term of an arithmetic sequence is of the form nr + s where r and s are two constants. So we can write a = (0)(r) + s, b = (1)(r) + s, c = (2)(r) + s, and so on. The sequence 5^a, 5^b, 5^c, ... becomes 5^(0r+s), 5^(1r+s), 5^(2r+s), ... where the nth term in the new sequence is given by 5^(nr+s).
We can then rewrite the terms of the sequence to obtain (5^s)((5^r)^0), (5^s)((5^r)^1), ..., (5^s)((5^r)^n), ... Note that the nth term of this sequence is of the form qp^n where q = 5^s and p = 5^r and therefore this is a geometric sequence (a geometric sequence is, by definition, one whose nth term takes the form qp^n for some constants p and q).