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 02-20-2011 #1 Lisasmith111   Join Date: Nov 2010 Posts: 36 Peter's GP Log question Here is something I found by poking around with Peter's Series. Question is why does it work (sort of!). ie 1/3 - 1/9 + 1/27... Suppose I give you a term from it say -1/6561. Now we know the common ratio could be calculated as follows:- term 2 / term 1 = -1/9 / 1/3 = -1/3 so common ratio = -1/3 we also know the first term = 1/3 Which means we could use the formula for finding a given term from the series ar^(n-1) Anywho here is the crux dear reader.... I could write -1/6561 = 1/3 * (-1/3)^(n-1) (Just to be clear in words the RHS is a third times minus a third raised to the power of n minus 1.) If I solve for n using logs I get -1 * log 1/6561 = log 1/3 + -1(n-1)log 1/3 -log 1/6561 = log 1/3 + (1-n)log 1/3 -log 1/6561 = log 1/3 + log 1/3 -nlog 1/3 (-log 1/6561 - 2log 1/3) / -log 1/3 = n = 10 (which is the wrong answer) However, If I do the following I get an answer which is closer to the truth:- So again I write -1/6561 = 1/3 * (-1/3)^(n-1) this time I multiply through by -1 and change the power around as follows:- 1/6561 = -1/3 * (1/3)^(1-n) log 1/6561 = -1.log 1/3 + (1-n)log 1/3 log 1/6561 = -log 1/3 +log 1/3 -nlog 1/3 log 1/6561 = -nlog 1/3 log 1/6561 / log 1/3 = -n n = -8 Now we can't have a negative term. So I could say we want the absolute value of n and therefore we get n = 8. Ie -1/6561 is the 8th term in the series. Which is correct. The trick seems to work but why does is work? (at least sort of!). But why? Is there a better way of doing it?