**Question Corner and Discussion Area**

How do we prove that 5^An arithmetic sequence is one for which the difference between consecutive terms is a constant, soa, 5^b, 5^c, . . . is a geometric sequence if it is known thata,b,c, . . . is an arithmetic sequence?

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 *n*th 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^(0*r*+*s*), 5^(1*r*+*s*), 5^(2*r*+*s*), . . . where the *n*th 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 *n*th 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 *n*th term takes the form *qp*^*n* for some constants
*p* and *q*).

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