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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.
Originally Posted by GrigOrig99 
