# Geometric Sequences and Series

• Aug 23rd 2011, 03:45 PM
GrigOrig99
Geometric Sequences and Series
Q. Show that if log a, log b and log c are three consecutive terms of an arithmetic sequence, then a, b and c are in geomtric sequence.

Eq.s: Un = a + (n - 1)d and Un = ar^n - 1

In an arithmetic sequence U2 - U1 = d
Therefore... (log b = a + nd) - (log a = a + nd - d)
b - a = d
now, log c - log b = (a + (n + 1)d) - (a + nd)
c - b = (a + nd + d) - (a + nd)
c - b = d
Thus, we see that the 3 terms of log a, b and c are all in arithmetic sequence.

In geometric sequence we have...
Un = ar^n - 1
so, log a = ar^n - 1, log b = ar^n and log c = ar^n + 1
if r = U2/ U1, then log a/ log b = r
log b/ log a = (ar^n/ ar^n - 1)
b/ a = 1/ -1 = -1
Also,
log c/ log b = (ar^n + 1/ ar^n)
c/ b = 1

Ans.: From text book: b^2 = ac, i.e. a geometric sequence

I'm confused as to how they have worked out the answer, as my solution, from the formula, involves the nth term of values. Can anyone help me out? Thank you.
• Aug 23rd 2011, 05:06 PM
Plato
Re: Geometric Sequences and Series
Quote:

Originally Posted by GrigOrig99
Q. Show that if log a, log b and log c are three consecutive terms of an arithmetic sequence, then a, b and c are in geomtric sequence.

Let's agree that we are mathematically mature and \$\displaystyle x=\log(a)\$
means that \$\displaystyle e^x=a\$.
If so then \$\displaystyle \log(b)=d+\log(a)~\&~log(c)=d+\log(b)=2d+\log(a)\$.
What does that mean?