Question Corner and Discussion Area
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).
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