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If Arithmetic and Geometric Sequences are simply Linear and Exponential functions respectively. Why then do we subtract the n variable by 1 when solving for certain terms in these sequences?


$$t_n=a\cdot r^{n-1}$$

I've tried exploring this question from a graphing perspective and it's clear this results in the sequence index starting at 1 rather then zero but is that really the only reason? Personally I'd rather start sequence indices at 0 as I've found going back and forth between linear equation techniques and sequence term formulas makes it easy to make off by 1 errors if one isn't careful.

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You have given one good reason why mathematicians like to start counting at $0$. A mathematical clock would have the hour labels $0$ to $11$. And $0$ would be on the extreme right of the clock, not on top. And the numbers would increase counterclockwise. – André Nicolas Jul 14 '12 at 18:58
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There's nothing wrong with using $n$ instead of $n-1$. It'll give you the same sequence, only shifted by 1. For example, suppose that we have two sequences $$ t_n=3(n-1)+2\qquad s_n=3n+2 $$ then we'll have \begin{array}{ccccccc} n & 1 & 2 & 3 & 4 & 5 &\dots\\ \hline t_n & 2 & 5 & 8 & 11 & 14 &\dots\\ s_n & 5 & 8 & 11 & 14 & 17 \end{array} There's no real difference between the two. Some sequences (sepecially those defined recursively) have properties that are more tidily defined when they start at a particular index, but as long as you're careful you can shift the terms in a sequence (or a series, for that matter) by any amount you wish.

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It is simply a matter of the indexing set you choose. Sometimes we choose $I=\{0,1,2,\dots\}=\Bbb N_0$ and sometimes $I=\{1,2,3,\dots\}=\Bbb N$. Using the former, we can write

$$\{t_n\}_{n \in I}\;\,/\;\,t_n=a\cdot r^n$$

and using the latter

$$\{t_n\}_{n \in I}\;\,/\;\,t_n=a\cdot r^{n-1}$$

There is no more mystery than that.

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