Find last term of geometric progression $1-4+16-\ldots$ if sum is $-209715$ How can we determine the last term of the series $$1-4+16-\ldots$$ if the sum is $-209715$? 
I can't seem to understand it. I've been lurking around the Internet for some help but can't seem to find any answer. 
 A: In a geometric progression $a,ar,ar^2,...$, the sum to $n$ terms is $\frac{a(r^n-1)}{r-1}$ and the $n$th term is $ar^{n-1}$.
Simply solve $\frac{1((-4)^n-1)}{-4-1}=-209715$. That will give you the number of terms you sum to, which turns out to be $10$, so that the last term is $1(-4)^{10-1}=-262144$.
A: This is an exercise in organized thinking.
To begin with, you are interested in a geometric progression. Recall that you know how to classify a geometric progression: by its first term, its ratio, and the number of terms it has.
To make things easier, we should assign those things names:


*

*$a$ is hereby defined to be the first term of the geometric progression

*$r$ is hereby defined to be the ratio of the geometric progression

*$n$ is hereby defined to be the number of terms of the geometric progression


Heck, let's even assign a name to the geometric progression itself: call it $G$.
Why did I do this? Because it's an automatic thing; once you're used to these sorts of problems, this comes naturally without even thinking about it. I didn't do any of this because I have a specific idea it will help solve the problem -- I did it because I know it's a way to help understand geometric sequences. (and therefore it will probably help understand the problem)

So that we've assigned names to things, let's look at the rest of the problem. You are given information about $G$:


*

*$G_1 = 1$ (The subscript $1$ means "the term in position number $1$")

*$G_2 = -4$

*$G_3 = 16$

*The sum of its terms is $-20975$


And we are looking for its last term.
Oh blah, I've had to use words again; words are hard to do arithmetic with. Let's assign names:


*

*$S$ is hereby defined to be the sum of the terms of $G$

*$L$ is hereby defined to be the last term of $G$


So we're told that $S = -20975$, and we're looking for $L$.
Now that we have names, we can try and start to understand things. What does it mean for $L$ to be the last term of $G$? That one's easy:
$$ L = G_n $$
So now, the problem is reduced to two questions:


*

*What do I know about the sum of a geometric progression?

*What do I know about finding particular terms of a geometric progression?


Well, it turns out that when I learned about geometric progressions, I learned formulas for both of these things! If you haven't learned the formulas, you are strongly suggested to look at your book again, or to seriously try to reason it out yourself -- or to ask another question on this site!
The formulas I learned are
$$G_k = a\, r^{k-1}
\qquad \qquad
S = \frac{r \, G_n - G_1}{r - 1}$$
(actually, I might have learned $S = a(r^n - 1)/(r-1)$, but this is closer to the way I think about it these days. They work out to the same thing, of course)
So now we have some equations we can work with:
$$a = 1
\qquad \qquad
a\, r = -4
\qquad \qquad
a\, r^2 = 16$$
$$ -209715 = \frac{a \, r^{n} - a}{r-1}
\qquad \qquad
L = a \, r^{n-1}$$
Now we've turned it into a simple algebra problem about solving equations. Solve it for $L$!
Note that you could have solved for $a$ and $r$ without even writing down equations -- you may even have been taught to do it this way. But experience has shown me that when I'm having trouble with a problem, writing everything out like this helps eliminate my confusion.

A note on conventions. In the above, I let $G_1$ be the first term. I actually prefer zero-up indexing -- that is, $G_0$ is the first term -- but I assumed you would be more comfortable with one-up indexing: that $G_1$ is the first term.

If you're clever or lucky, you might have found a shortcut to solving the equations; if I had thought slightly differently, I would have written them as
$$a = 1
\qquad \qquad
a\, r = -4
\qquad \qquad
a\, r^2 = 16$$
$$ -209715 = \frac{r L - a}{r-1}
\qquad \qquad
L = a \, r^{n-1}$$
This form is somewhat easier to solve for $L$!
A: Just find the biggest power of 4 that is greater than 209715. Then negate it.
