algebra precalculus - Finding which term in a sequence the last term of a sum corresponds to.

To sum the first $n$ terms of a geometric sequence we use the formula,

$$S=\frac{a(1-r^n)}{1-r},$$

where $a$ is the first term of the sequence, and $r$ is the common ratio.

Now consider the geometric sum of an analogous geometric sequence,

$$1+\frac12+\frac14+\frac18+...+\frac1{128}$$

Clearly in this case, $a=1$ and $r=\frac12$, but what about $n$? After some trial and error you can see that $\frac1{128}$ is the eighth term of the corresponding geometric series, but is there a systematic (formulaic) way to determine what term in the sequence the last term in the finite sum corresponds to?

Answer

In a geometric series sum, the first term is $a,$ the next is $ar,$ the next is $ar^2,$ and so on.

So if you let the last term be b, we have that $\frac{b}{a}=r^k$ for some k. Since we can find the common ratio by dividing the second term by the first, it shouldn't be hard to find k. So the last term is the $(k+1)$th term (as the first time is $ar^0$).

With the series above as an example, we have $a=1$ and $r=\frac{1}{2}$ as mentioned, and since the last term is $b=\frac{1}{128}$ we have $k=\log_{\frac{1}{2}}\frac{1}{128}=7$, so the last term is the eighth term.