SOLUTION: For the geometric sequence {{{a}}},{{{ ar}}},{{{ar^2}}} ... (to infinity), show that the sequence {{{loga}}}, {{{log((ar))}}}, {{{log((ar^2))}}}....(to infinity) is an arithmetic s

Algebra ->  Sequences-and-series -> SOLUTION: For the geometric sequence {{{a}}},{{{ ar}}},{{{ar^2}}} ... (to infinity), show that the sequence {{{loga}}}, {{{log((ar))}}}, {{{log((ar^2))}}}....(to infinity) is an arithmetic s      Log On


   

Question 1146288: For the geometric sequence a,+ar,ar%5E2 ... (to infinity), show that the sequence loga, log%28%28ar%29%29, log%28%28ar%5E2%29%29....(to infinity) is an arithmetic sequence.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Two consecutive terms of the given sequence are ar^n and ar^(n+1).

We need to show that the difference between log(ar^n) and log(ar^(n+1)) is a constant.

log%28%28ar%5E%28n%2B1%29%29%29-log%28%28ar%5En%29%29 =
=
log%28%28r%29%29

r is the constant ratio between terms, so log(r) is a constant.