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Since it's almost the new year, this image (click the image for a larger view) is making the rounds on facebook again, and everyone is reading it and saying, "WHAT? How does that work out to $667.95?"

The answer is, it doesn't. It works out to *exactly *what you thought it would work out to: 365 pennies, or $3.65. Unless it was Leap Year, and then it would work out to $3.66. Not terribly hard math.

*However*, even though "a penny a day" is what they said, "a penny a day" is not what they *meant.* How do I know that? I'll show you in just a minute.

What they meant was, on January 1st you put in one penny. On January 2nd, you put in two pennies. On the 3rd, three pennies. And so on, until December 31st, when you put in 365 pennies, because it's the 365th day of the year.

How do I know that's what they meant? Because there's a really nice formula to calculate the sum of the first n integers:

Sum = n(n + 1)/2

In this case, n = 365, so the sum is:

Sum = 365(365 + 1)/2 = 66,795.

And that's how I know that's what they meant - because if that's what they meant, the math works out correctly to $667.95.

And by the way...if it was leap year?

Sum = 366(366 + 1)/2 = 67,161, or $671.61.

So yes, this will work out to save you a pretty good sum of money by the end of the year. Maybe enough to pay a month's rent, depending on where you live!

Of course, it's not as good as *doubling *the number of pennies each day; that would save you a *boatload* of money over the course of a year:

1 + 2 + 4 + ... + 2^{n} = 2^{n + 1} - 1

So: 1 + 2 + 4 + ... + 2^{365} = 2^{366 } - 1 = 1.5 x 10^{110}.*

Actually, scratch my last comment; that's not a boatload of money. If a penny's volume is 0.35 cm^{3}, or 0.00000035 m^{3}, that's a volume of 5.25 x 10^{103} m^{3}. Considering the volume of the sun is 1.4 x 10^{27} m^{3}, I don't think you're going to fit those pennies in a jar, a boat, or even all the planets of the solar system. Maybe we should just stick to the original plan.

Incidentally, there's also a "Dollar a Week Challenge" in which you save one dollar the first week, two dollars the second week, and so on for an entire year. It sounds like a lot more, since you're saving a dollar instead of a penny, but there are only 52 weeks in the year, so it works out like this:

Sum = 52(52 + 1)/2 = $1,378, which is only a litte more than twice the penny challenge.

If you wanted to get really interesting, you could do a "Dollar Square Weekly Challenge," which would look something like this:

In the first week, you save 1^{2} = 1 dollar.

In the second week, you save 2^{2} = 4 dollars.

And so forth. It turns out, there's a nice formula for the sum of squares as well:

Sum = n(n + 1)(2n + 1)/6

Sum = 52(52 + 1)(2 x 52 + 1)/6 = $48,230. That's a lot of money! How would it compare to a Penny Square Daily Challenge?

Sum = 365(365 + 1)(2 x 365 + 1)/6 = 16,275,715 pennies, or $162,757.15. So even though the Dollar Challenge saves more than the Penny Challenge, the Penny Square Challenge saves more than the Dollar Square Challenge.

It would be interesting to see how many weeks you'd have to do the Dollar Square Challenge in order for it to surpass the Penny Square Challenge for a year. I'll leave that do the reader to figure out!

* This formula is actually for a leap year. Since the first term of the sequence is 2^{0}, the 366th term is 2^{365}.

Emmanuel from Papua New Guinea asks, "How do find the common ratio of a geometric sequence if the ratio of the fourth and second term are given?"

Well, Emmanue, the short answer is: you can't!

Let's suppose the second term of geometric is 4, and the fourth term is 16. You might think, "Oh, that's easy - the ratio must be 2, because 4 x 2 is 8, and 8 x 2 = 16!" But that's not necessarily true - maybe the ratio is -2! 4 x -2 = -8, and -8 x -2 = 16.

The problem is, in a geometric sequence, all the even-numbered terms will have the same sign, but that won't tell us anything about the sign of the odd-numbered terms, and that information is needed to find the ratio. But we *can* set up an equation that'll give us the possible values.

Let's say the second term is 2, and the fourth term is 18. Then

ar = 2, and ar^{3} = 18

If we rewrite the second equation as ar(r^{2}) = 18 we can subsitute the first equation in place of ar, giving:

2r^{2} = 18, or

r^{2} = 9.

Now, it's tempting at this point to say, if r squared is 9, then r must be 3, but you're missing a possibility if you do that, because 9 has two square roots: 3 and -3. These are your two possible ratios. We don't know what the ratio is, but at least we've narrowed it down to two possibilities!

By the way, as a side note, in order to get my students to avoid missing a solution in an equation like r^{2} = 9, I tell them they have to solve the equation like this:

r^{2} = 9

r^{2} - 9 = 0

(r - 3)(r + 3) = 0

Therefore r - 3 = 0 or r + 3 = 0, which leads to r =3 or r = -3.

It's more work, but it keeps them (most of the time) from forgetting a root!

We've had several questions recently about how to find terms of a geometric sequence, if you've been given specific information about the sequence, such as, what two of the terms are.

So let's consider a couple examples. For starters, let's say we have a sequence in which the second term is 4, and the fifth term is 27/2. How do we find the first term and the common ratio?

First, since the n^{th} term is ar^{n-1}, we know that ar = 4, and ar^{4} = 27/2. So what do we do with these two pieces of information? Well, we can take the second equation and divide it by the first one, to obtain r^{3} = 27/8, or r = 3/2.

Since we know that r = 3/2, we can plug that into our first equation, and get (3/2)a = 4, so a = 8/3.

Not too bad, right?

So let's try another one that works out to be just a little different.

If the second term is 24, and the fourth term is 6, find the first term and the common ratio.

We start out the same way:

ar = 24, and ar^{2} = 6. Dividing these two equations gives r^{2} = 1/4.

At this point, it's tempting to take the square root of both sides and say r = 1/2. (I see algebra students do this all too often!)

However, it's important to remember that 1/4 has TWO square roots: 1/2 and -1/2.

This means there are TWO possibilities for a. If r = 1/2, then a = 48, but if r = -1/2, then a = -48.

As a general rule of thumb, if you're given two terms of the same parity (both odd numbered terms, or both even numbered terms) you're going to have two possible solutions.

Thanks for asking, and good luck with your sequences!

Professor Puzzler

Karylle from Marinduque wants to know how you can find the ratio in an infinite geometric series, if you know that the sum is a particular multiple of the first term.

This is in that interesting class of problems in which you feel like you don't have enough information to solve it. After all, there are a lot of unknowns (the first term, the ratio, and the sum) and you'll only have one equation to work with. Whenever my students have problems like this, I tell them to simplify the equation as much as possible, and see what happens!

So let's take an example. Let's say that you know the sum of the infinite series is 5 times the first term. Can you find the common ratio?

Well, the sum of the series is a/(1 - r), and that is 5a.

a/(1 - r) - 5a = 0

Ah...I see what's going to happen already - the a is going to factor out:

a(1/(1 - r) - 5) = 0

so either a = 0 (in which case, r could be anything, right? It's a really boring series, with all the terms equal to zero, but hey, it works!) or 1/(1 - r) = 5. This leads to:

5 - 5r = 1

5r = 4

r = 4/5

I would be inclined to say that the problem should be reworded to state that the first term is not 0, in order to avoid having "r can be any real number" as the answer. If you add in that proviso, then the answer is r = 4/5.

Hope that helps, Karylle!

Professor Puzzler

Jana from the Philippines wants to know, if you know the second and fourth terms, how do you find the first term?

Well Jana, that's a bit of a trick question, because if you don't tell me what KIND of sequence it is, I can't figure out the first term (or any other terms for that matter!). The most common types of sequences you might be talking about are: arithmetic and geometric. Of course, that's not all the possiblities. It could be a Fibonacci sequence, for example, or even a random list of numbers. Of course, if the sequence is random, there's really no way we can figure out any terms, is there?

So let's suppose the sequence is either arithmetic or geometric. And let's use the same numbers for both types of sequences. We'll say that the first term is 12, and the fourth term is 48.

### Arithmetic

The nth term of an arithmetic sequence is given by a+(n-1)d, where a is the first term, and d is the common difference.

Thus, 12 = a + d, and 48 = a + 3d. This is a system of two equations in two unknowns, and if we solve it, we find that d = 18, and therefore, a = -6. That's our first term.

Geometric

Then nth term of a geometric sequence is given by ar^(n -1). Thus, 12 = ar, and 48 = ar^3. If we divide the second equation by the first one (which we can only do if a and r are not zero!) we obtain 4 = r^2, from which r is either 2 or -2. Thus, the first term is 6 or -6.

### Just for Kicks

A Fibonacci sequence is a sequence in which each term is the sum of the two previous terms. So if this was a Fibonacci sequence, the third term would have to be 36, since 12 + 36 = 48. And this leads to the first term being 24, since 24 + 12 = 36.

There you go! How to find the first term, given the 2nd and 4th terms, for three types of sequences.

Good luck!

Professor Puzzler