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Eleventh grader Kausar asks, "I had this question several times but i do not know how to do it. You are given the 5th and 7th terms of a geometric sequence. Can you determine the 29th term without finding the general term?"
Hi Kausar, the answer to your question is, "Yes, you can." Having said that, I'd like to add that if you'd been asked for the 28th term, or the 30th term, the answer would be "No, you can't." Let's take a look at why.
Let's say the fifth term is 32, and the seventh term is 8. Then we can reason as follows:
"The sixth term is the fifth term times the common ratio, and the seventh term is the sixth term times the common ratio. Therefore, the seventh term is the fifth term times the common ratio squared."
Algebraically, this reasoning looks like this:
a6 = a5r; a7 = a6r; a7 = (a5r)r = a5r2. Thus, 8 = 32r2, which leads to r2 = 1/4.
Now, without bothering to get the general term, we ask, "How many times would I have to multiply the seventh term by r2 to get the 29th term? And the answer is: (29 - 7)/2 = 11.
So a29 = a7r11 = 8(1/4)11, which is approximately 1.9x10-6.
So why does this work for the 29th term, but not the 28th? Because (28 - 7)/2 is not an integer, so we can't get there by multiplying by r2. We'd have to find r, and multiply by r (28 - 7) times.
So why can't we do that? Well...because the equation r2 = 1/4 doesn't have just one solution; it has two. r = 1/2 or r = -1/2. So you have two possible values for the 28th term.
I occasionally get hired to write problems for math competititions, and one of the "tricks" I'll occasionally pull on students is to give a sequence like this, where students are likely to end up with an equation like r2 = 1/4, and then forget that this has two solutions instead of just one. It's amazing how often a simple trick like that will trip students up!
Twelfth grader Jodi asks, "Hi Professor, if you could help me understand this problem I would highly appreciate it. Problem: An ant starts digging a tunnel. On the 1st day he digs 9m north. The 2nd day he digs 6 m east. The 3rd day he digs 4m south. He continues to dig in this pattern. Question 1: How far will he dig in total? Question 2: How far north from his starting point will he end up?"
Hi Jodi, this is definitely a fun problem. I'm going to answer these two questions, and then I'm going to pose two more questions for you to think about.
Question 1: How far will the ant dig?
Each day the ant digs 2/3 of what he dug the day before. Thus, his total distance is
Eq. 1: D = 9 + 6 + 4 + 8/3 + 16/9 + ...
Now, there are formulas you can use to calculate that sum, and if you're in a Pre-Calculus class (or higher) you probably know those formulas. I'm going to show you how to get that sum without a formula. I'm going to take the equation and multiply it by 2/3:
Eq. 2: 2/3D = 6 + 4 + 8/3 + 16/9 + 32/27 + ...
Now, if I subtract the second equation from the first one, all but one of the terms on the right hand side of the equation cancel out, and I get the following:
Eq. 3: D/3 = 9
This leads to D = 27, which is the answer to Question 1. If you want to make sure, pull out a calculator and punch in a bunch of these terms; you'll see that it doesn't take more than 7 or 8 terms for you to start getting close to 27.
Question 2: How far north from his starting point will he end up?
Before we get started, I'd like to point out that the problem seems just a tiny bit ambiguous. There are a couple ways I could read this:
- The ant digs North, then East, then South, then repeats this pattern, going North, East, South, North, East, South, etc.
- The ant digs North, then East, then South, then North, then West (completing the compass pattern) before repeating all four directions.
Strictly speaking, interpretation #1 is the precise reading of the problem, but I have a feeling interpretation #2 was intended, so I'm going to go with that one.
To solve this, we treat a distance south as a negative distance, and we look at the odd numbered terms of the original sequence (because the even numbered terms are east/west motions, and don't affect the north/south distance).
Eq 4: DN = 9 - 4 + 16/9 - 64/81 + ...
This time the common ratio is negative (we know this because the terms are alternating signs). It is -4/9. So you can do the process I did above, multiplying the equation by -4/9 and then subtracting. Or you can use the formula I referred to above, but didn't tell you:
Eq 5: S = a/(1 - r), where a is the first term and r is the ratio.
DN = 9/(1 - (-4/9)) = 9/(13/9) = 81/13
Question 3: What about the other digging pattern?
We made the assumption that the problem intended for the ant to dig in all four compass directions, but that may not have been the intent. What if the western direction is skipped each time? How does that affect the answer. It definitely makes it more interesting; I think you'll end up needing to treat it as two separate series: DN and DS, and then find the difference.
Question 4: What's the total displacement?
Displacement is a word we use a lot in Physics. It's different from the concept of distance (which is how far the ant traveled). Displacement is how far (and in what direction) the ant ended up from where he started, and it's a combination of the distance north, and the distance east. To find that answer, you would need to set up a DE equation (distance east), and then do the Pythagorean Theorem on DN and DE to find the hypotenuse, which is the distance from his starting point to his ending point.
Since displacement is a vector (magnitude and direction) you'd also have to do a bit of trig to find the direction in which the ant ended up from his starting point.
Those are problems I'll leave you to contemplate. Thanks for asking!
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 + ... + 2n = 2n + 1 - 1
So: 1 + 2 + 4 + ... + 2365 = 2366 - 1 = 1.5 x 10110.*
Actually, scratch my last comment; that's not a boatload of money. If a penny's volume is 0.35 cm3, or 0.00000035 m3, that's a volume of 5.25 x 10103 m3. Considering the volume of the sun is 1.4 x 1027 m3, 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 12 = 1 dollar.
In the second week, you save 22 = 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 20, the 366th term is 2365.
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 ar3 = 18
If we rewrite the second equation as ar(r2) = 18 we can subsitute the first equation in place of ar, giving:
2r2 = 18, or
r2 = 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 r2 = 9, I tell them they have to solve the equation like this:
r2 = 9
r2 - 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 nth term is arn-1, we know that ar = 4, and ar4 = 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 r3 = 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 ar2 = 6. Dividing these two equations gives r2 = 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!
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