Problem number 21. What is the measure of the acute angle of a rhombus with side of length equal to the geometric mean of its diagonals? So to solve this problem, it sort of requires you to know a few definitions and a few formulas. But if you know them, the problem is fairly straightforward. If you don't know them, it's kind of difficult to find a way to go forward. Now the first thing we need to know is the definition of a geometric mean as opposed to a regular mean. What's the difference? So the difference is that the geometric mean can be thought of as the nth root of a product of n numbers. So if you're taking the product of two numbers, you're taking the square root of that product. And in this case, since there are two diagonals, we're going to take the square root of the product of their lengths. And that's going to be equal to the side length. So we have A equals PQ, the square root of PQ, where A is the side length and PQ are the diagonals. Or A squared is equal to PQ. So again, this is the side length. These are the two diagonal lengths. And now the other thing you need to know is the two different, two useful formulas for the area of a rhombus. There are three different methods for finding the area of a rhombus that are sort of well-established, and two of them involve using these two formulas right here. The first one being 1 half times the two diagonals, and the other one being the side length squared times the sine of the acute angle, which we will call X. Now we're trying to solve for X here, and in particular, once we compare these two formulas for the area, we note the fact that A squared is equal to P times Q, based on what's given in the problem, since this is the geometric mean statement given in the problem. And if A squared and PQ are equal to each other, then that means that the things that we're multiplying them by must also be equal to each other, since these effectively cancel each other out. So that means 1 half is equal to sine of X, and the sine of what angle is equal to 1 half? Well, that turns out to be 30 degrees, and you just know that from your knowledge of the unit circle. So our final answer is B. Problem number 22. Which of the following is a graph of the function Y is equal to the square root of the absolute value of 1 plus X times 1 minus absolute value of X? Now looking at these answer choices, we find that it's really, there's no special method to going about solving this problem. You want to take a look at the behavior of the graph at various points, and sort of see if it should feel more like a curve, or if it should feel like a straight line at various points. So in particular, we want to take a look at this closer area in the center first, and decide, okay, should the positive side be sloping in a straight line, or should it be on a curve like this? Now starting from the positive side, we can go ahead and plug in some positive values for X. Let's say we could plug in 3 4ths for X. In that case, we get 1 and 3 4ths, which is 7 4ths times 1 minus 3 4ths, which is 1 4th. So it's 7 4ths times 1 4th, which is equal to 7 over 16, or square root of 7 over 4, when we take the square root. Now, how helpful is this? Let's go ahead and compare it to another point. Let's suppose we have 1 half as X. Well in that case, we have 3 halves times 1 half, which is equal to 3 over 4, which gives us square root of 3 over 2. Now these numbers, square root of 7 over 4, square root of 3 over 2, these aren't particularly nice numbers, and it's not easy to imagine there being sort of a slope between them of a straight line. Now it seems a little bit hand wavy, but if you're taking an actual exam, this is one way that you could sort of get a feel for how the graph should go. Now just looking at the positive side, there's no clear pattern of, okay, if you want to get from, if you want to go this distance on X, you have to go this distance by Y. It doesn't seem like there's a linear slope to it. But we don't know anything for sure. So let's go ahead and look at the negatives and see if that is more obviously a slope. So if we let X equal negative 1 half, for instance, well, then we have 1 half times 1 half, which equals 1 fourth, and the square root of that is equal to 1 half. So the point negative 1 half for X gives us 1 half for Y. Now that right away seems like it should be giving us one of these straight lines on the negative side, because those have a slope of 1, these small lines here. And so that would line up the negative 1 half and 1 half values for X perfectly. Now to test this, we can go ahead and name another nice X value. Let's say negative 1 fourth, and then we would have 3 fourths times 1 minus absolute value of negative 3 fourths, which is also 1 fourth. So we have 1 fourth times 1 fourth is equal to 1 over 16, and the square root of that is equal to 1 over 4. So again, we have this point that is consistent with a line that is of slope 1. And if you compare the two different lines, it definitely feels like the negative side has a very obviously linear slope, and the positive side doesn't seem to have a slope like that. So just by that reasoning alone, we look at the different options, and we see that there's only one where on the negative side there's a linear slope, and on the positive side there's a curve to it. So we conclude that our final answer is D. Problem number 23. Which of the following numbers cannot be the value of X plus square root of X, where X is an integer? Now looking at our answer choices, we see that all of them are integers as well, and that means that we want square root of X to also be an integer. We can't have some sort of irrational number for square root of X because it's not consistent with any of our answer choices, which means that X must be a perfect square. So what we can do is we can look at the answer choices and see if we can come up with an X that works. If we can, then we can eliminate the answer. If we can't, then we leave it on the side and maybe think about it again after cycling through all the answer choices. Notice, by the way, that 870 is a pretty big number to consider because we don't really have any perfect squares just lying off the tops of our heads that necessarily work for large numbers like this. But if we're looking at some of these smaller numbers, then maybe we can sort of try imagining some numbers that we already know are perfect squares. So for instance, in the case of 110, well, in this case, we can let X equal 100, and the square root of that is 10. So 100 plus 10 gives us 110, and that means that 110 is a possible value, so we have to eliminate it. Now answer choice C, 90. If we let X equal 81, well, then we have 81 plus 9, which does equal 90, so we can eliminate that. And if we go ahead and check 30, well, the closest perfect square below 30 that we can think of is 25, and 25 plus its square root, which is 5, is certainly equal to 30, so we can eliminate that. So we've already narrowed it down to 870 and 60 just by trying to sort of come up with examples of what X can be. Now, if we can come up with an X that works for 60, then we can eliminate it, and we know that our final answer is A. Or we can realize that there is no perfect square that works for 60. So the nice thing about it being 60 is that we know all of the perfect squares that are less than 60, so we definitely don't need to check 870. Now, the first perfect square that's below 60 is 49, and that's 7 times 7. 49 plus 7 is definitely not equal to 60, so that's not going to work for an X. 64 is bigger than 60, so we're not going to consider that, and the next perfect square would be 6 times 6, which is 36, and 36 plus 6 is way too small to be 60, so none of the perfect squares that are near 60 sort of work. So we can conclude that our final answer should be D. Problem number 24. If f of X equals 2x over 3x plus 4, and f of g of X is equal to x, then g of X is equal to what? So, looking at the problem, we see that f of g of X is equal to x, and by definition, that means that f and g are inverse functions of each other. So our goal is to find the inverse of f of X, and that's going to be equal to g of X. Now, if you don't recall the method for finding an inverse of a function, it's pretty straightforward. You start by... Okay, so normally you imagine that f of X is equal to y, so y equals 2x over 3x plus 4. Now, if you want to find the inverse, just flip the y's and the x's. So instead of y equals 2x over 3x plus 4, imagine X equals 2y over 3y plus 4, and then you just try solving for y, and that's going to give you g of X. So solving for this, we start first by trying to find a way to make this fraction nicer. Ideally, we would like to break this fraction up into two parts, so it's more manageable, because in the end we want to isolate a y on one side of the equation. So if we take the inverse of both sides, we have 3y plus 4 over 2y equals 1 over x, and this fraction now breaks up into 3 over 2 plus 2 over y. And this is equal to 1 over x. Subtracting 3 halves from both sides of the equation, we get 2 over y equals 1 over x minus 3 halves, and we can take both of these fractions and alter them so that they have the same denominator, multiplying this by 2 over 2 and multiplying this by x over x, and that gives us the new fraction 2 minus 3x over 2x, and this is equal to 2 over y. Now we want a y in the numerator, so we just take the inverses again, and we get y over 2 is equal to 2x over 2 minus 3x. To isolate the y, we need to multiply both sides of the equation by 2, so we get y is equal to 4x over 2 minus 3x. And looking at the answer choices, that's consistent with answer choice D, it looks like. So that's going to be our final answer. Problem number 25. For how many real numbers a does the equation x squared plus ax plus 2007 equals 0 have two integer solutions? So we want to imagine all of the different ways that we could pick an a so that x squared plus ax plus 2007 can factor into two parts that each have an integer term. Now this basically depends on knowing how we can factor 2007, because we would take two of the factors of 2007 that multiply to 2007 and add them, and that would give us a potential a value. Now the tricky part about this is we don't know how 2007 necessarily factors. It's not like a number like 24 or 20 or something where you can very quickly just sort of figure out what the factors are by listing all of them. Instead, it makes sense to start by trying to figure out the prime factorization of 2007 and see what happens next, just sort of feeling out for what factors there might be. You realize that 2007 is divisible by 3, and in fact by 9, since the sum of its digits is equal to 9. And once you divide 2007 by 9, you get 223, and 223 happens to be a prime number. You can run it through all the tests. It's not divisible by 2, by 3, by 5, or by any of the other numbers. It's not divisible by 7. A quick way to check that is it's equal to 210 plus 13. 210 is divisible by 7, but 13 is not, so the sum is not divisible by 7. Going through it, the problem seems like it might be a little bit unreasonable if 223 did have any more factors that were, for instance, double digit numbers. So, looking at it, 223 probably doesn't have any more factors. In fact, 223 is a prime number. So we know that 2007 is equal to 223 times 3 times 3, and so that means we can factor this 2007 equation into one of three things. It can be x plus or minus 1 times x plus or minus 2007, x plus or minus 223 times x plus or minus 9, or x plus or minus 669 times x plus or minus 3. Now keep in mind, if we have a plus here, we need to have a plus here since we need a positive times a positive, or we need a negative times a negative since it's a positive 2007 here. If one of these terms would be negative and the other would be positive, we'd have a negative 2007 here. So we don't even need to know the a values, although we could briefly check just to make sure that there are no, there's no overlap or anything, but that shouldn't happen mathematically. So without even knowing the a values, we can go ahead and see that we have plus or minus 1 times plus or minus 2007. So there's two a values here. It's going to be 2008 and 2006, and we also have two values from this factoring, since it's plus or minus and plus or minus, and two values from this factoring. So we have a total of six different ways to factor the equation depending on how we pick a. So our final answer is C. Problem number 26. What is the sum of this series? So looking at the series, we see that it would be pretty much impossible to add all of the numbers up by brute force, unless there's some sort of nice trick or pattern or something. So we should be looking for a pattern. Now generally when I have a rational number in the numerator and some sort of square roots in the denominator, my first instinct is to go ahead and rationalize the denominator, and the way we go about doing that is by multiplying the denominator by its conjugate, which is basically the same number, but with a negative instead of a positive. So if you're taking the sum of these two numbers, then you're taking the difference instead. So let's start by looking at just the first two terms and seeing how they add up. Well, we're going to go ahead and multiply this by 2 times square root of 1 minus square root of 2 over 2 times square root of 1 minus square root of 2, and we're going to multiply this by 3 times square root of 2 minus 2 square root of 3 over 3 square root of 2 minus 2 square root of 3. And that's going to give us 2 minus square root of 2 in the numerator over 4 minus 2. That's what happens when you multiply out the two denominator terms, plus 3 square root of 2 minus 2 square root of 3 over 18 minus 12. That's what happens when you multiply this denominator by its conjugate, and if we simplify this further we get 2 minus square root of 2 over 2 plus 3 square root of 2 minus 2 square root of 3 over 6. And now notice here, so we have a 2 over 2 here, which is equal to 1, and we're subtracting square root of 2 over 2. And then here we have an addition of 3 square root of 2 over 6, but the 3 and the 6 reduce down to 1 over 2. So in fact, this is just a positive square root of 2 over 2, which cancels out this negative square root of 2 over 2. So this term will cancel out this term, and now we see that there's maybe room for a pattern. And in fact, if you go ahead and were to include the following term, which would be 1 over 4 square root of 3 plus 3 square root of 4, and you would go ahead and rationalize the denominator the same way as we did for the first two terms, then we would end up getting the first part of the term canceling out with negative 2 square root of 3 over 6, and then the next term standing alone and canceling out with something from the next term. So basically the pattern is this very first term doesn't cancel out with everything. Everything in the middle does cancel out, and then the very last term, which would be in this case, whatever we get at the end of after rationalizing this, so 100 times square root of 99 minus 99 square root of 100 over the product of those two things here. So writing that out since that's a little hard to follow with words, and these terms are canceling out here, we end up getting 2 over 2 minus 99 times square root of 100 over 9900. 9900 is what you get when you multiply this denominator by its conjugate, and this 99 times square root of 100 is the second term of... It's basically coming from this 100 square root 99 minus 99 square root of 100. The minus 99 square root of 100 in the numerator is this, and then again the product of the two conjugates is this, and if we were to go ahead and just simplify both of these fractions, we would end up getting 1 minus 1 over 10, where 99s cancel out. So we have square root of 100 over 100, but square root of 100 is equal to 10, so it's 10 over 100, which is equal 1 over 10. So we end up getting 1 minus 1 tenth, which is equal to 9 tenths. So our final answer is C. Problem number 27. A group of five friends plans to exchange presents so that each person will give only one present and receive only one present. In how many ways can this be done, assuming that no one will give a present to himself or herself? So, I'm gonna go ahead and list the math first, and then explain where all of the numbers are coming from. So in particular, let's call the five people A, B, C, D, and E, just to make this easier to keep track of. Now, the words are going to be a little bit difficult to follow, so try to listen closely and sort of keep track of all of the different possibilities. Now, it doesn't matter who we start with, who gives the first present. The order that they give the presents in doesn't matter. But we'll go ahead and just assume that A is giving the first present. Now, A has four different choices, B, C, D, or E. Now, it doesn't matter who A gives it to. Let's just go ahead and assume for the sake of argument that B is going to get the present from A. The same argument would hold if it were C, D, or E, so don't worry about them. So we have four choices for A, and now B has this following choice. It can give a present to A, or B can give a present to C, D, or E. It's basically either give it to A, the person who already gave a present, or give it to C, D, or E, who haven't decided where their presents are going yet. So let's assume after A giving the present to B, that B gives the present to A. Now, if we were considering A giving it to C, then we would be thinking about C giving it to A right now. So it's basically, what if these two people just give each other presents? What happens then? Well, A had four choices, and we're looking at B giving a present to A, and in that case, we have C, D, and E remaining without presents. So it doesn't matter who goes next. Let's just suppose it's C. Now, C has two choices. He can give a present to D or E, and whoever that person is, let's say it's D, they cannot give a present back to C, because then E is left without anyone to give a present to. They cannot give a present to themselves, so they must give a present to E. So there's only one choice here. So if A gives a present to B, where there are four choices, and then B decides to give a present to A, where it's just one choice, because we're considering this. We'll consider this separately. So it's four choices times one choice, and then we have two choices, so C can give to either D or E, and then whoever C gives the present to, whether it's D or E, only has one choice to give it to whoever hasn't received a present yet or who hasn't given a present yet. So overall, that's gonna be four times one times two times one, or just four times two times one, and that is how many possibilities we have when A gives to someone, and then that someone gives right back to A. That's what we considered here. Now, what if A gives to someone, and then that someone gives their present to someone other than A? So what if A gives a present to B, and then B gives a present to C, D, or E? Well, in that case, A has four choices, and this next person has three choices, so B has C, D, or E. Now, let's suppose that you just pick C. As usual, the same argument holds whoever you go for. So now C can pick either A, D, or E to give a present to. That's three choices, and none of them lead to a contradiction, and in this case, C can give to A, D, or E, and whoever they pick, the other two people, or maybe not the other two people, but there's not going to be one person left alone who's not giving or receiving a gift. There's not going to be some issue like that, and you can go ahead and check that for yourself, but C can give a present to A, D, or E, and there won't be a problem here. So there's three choices here. Let's suppose that C gives a choice to A. In fact, let's go ahead and look at all three of the cases. If C decides to give the present to A, well, then A gave a present, B gave a present, and C gave a present. They all just cycled around each other, so there's only one remaining way for the presents to go. D and E have to swap, so that's just one choice left. So that would give us four times three times three times one. But okay, let's suppose that A gives a present to B, B gives a present to C, and C doesn't give one to A, but C gives one to D or E. Let's suppose D. Well, in that case, D has to pick who to give the present to, either E or A. Well, if D gives the present to A, then E is left all alone. E can't give or receive a present from anyone, so this doesn't work. So in fact, D must give a present to E. So recapping, if A gives a present to B, B gives a present to C, and C gives a present to D, then D must give a present to E. If he gives a present to A, then E is left all alone. So we have A having four choices, B, C, D, or E. B having three choices, or rather, B having three choices, C, D, or E. And then whoever gets it also has three choices, A, and then the other two, and then after that, there will only be one possibility remaining. So it'll be four times three times three times one. Now, it's a little bit difficult to keep track of all these possibilities at first, especially because what we're doing, we're breaking this up into cases where we're assuming that A gives a present to someone, and then that someone will either give the present back to the original giver, or give it to somebody who hasn't given a present yet. But once we count up all of these possibilities using this reasoning, then we get these two products, which are equal to eight and 36 respectively, and their sum is equal to 44. So our final answer is C. Problem number 28. The sequence given by one, two, three, four, five, one, two, three, four, five, one, two, three, four, five, and so on, is written in an infinite grid as shown. What number will be in the square line, 100 squares above the shaded square? So looking at this problem, and counting off how many squares we have to get to the square above the original one, and then the square above that, and then the square above that, well, we'll start by counting that and seeing if there's any sort of pattern. So counting the number of moves it takes, so this being one move, one, two, three, four, five, six, seven. It takes seven moves to get to the square above the first one. And then there's one, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15. So it's 15 to get to the second one. And in fact, if you were to count once again, you would see that it takes 23 to get to the third square. So seven, 15, 23, what's the pattern? Well, the pattern is that it's going to keep on adding the same amount over and over again. Our first number is seven, but then seven plus eight is 15, and then 15 plus eight is 23. So the next square up above the previous square is always going to be an additional eight squares on top of however many squares before there was. So it forms an arithmetic sequence where the original number is seven, and then each following number is eight greater than the previous number. So we can use this, and we can use the fact that it's a finite arithmetic sequence. There are 100 terms in the sequence, and we can go ahead and find the sum of these 100 terms to figure out how many squares we are actually moving along in order to get to the square that's 100 above square number one. So if there are 100 such terms, what we can do is we can take the very first term, which is seven, and the very last term, which is seven plus 99 times eight. And seven plus 99 times eight plus an additional seven is equal to 806. So when we pair up the first and last term and add them up, we get 806. Now, if we were to go ahead and pair up the second and 99th term in the same way, we would find that that sum would also be equal to 806 because this is an arithmetic sequence. And then the third and 98th term also sum up to 806, and the fourth and 97th term also pair up as such. So using this sort of pattern, we can just take the 50 pairs of numbers that we will get, pairing the first 50 terms with the second 50 terms, and we get 806 times 50 is equal to 40,300. And in fact, this is a common method for finding a large sum of a finite arithmetic sequence because the corresponding terms, like the outermost terms and then the second from the outermost terms and the third from outermost terms, in an arithmetic sequence, those always add up to the same sum. So using this sort of a pattern is a very efficient way of figuring out the sum of a large sequence. Now, seeing that the sum of this particular sequence is 40,300, notice that this number is clearly divisible by five since it ends in a zero. Now, the pattern is repeating the numbers one through five over and over again. So starting with the first square, we have one, two, three, four, five moves, and we're back to a square with a one on it. And again, one, two, three, four, five. And again, we're back to a square with a one on it. So because this number is a multiple of five, we're essentially going through the five numbers over and over and over again a whole bunch of times. And at the end of it, we're going to go right back to square one. If this were, for instance, 40,303, then we would go ahead and be able to sort of skip the first 40,300 squares and then just make three moves. So one, two, three, and we would know that that would be square labeled number four. In particular, there's a name for this sort of approach. You can think of it as something very similar to counting the time on a clock, how you go through the numbers one through 12 or one through 24, and then you start all over again with the number one. This is called modular arithmetic, and it's not very hard to understand. You can think of it as just adding up to a certain number and then it resets to zero. So for instance, right now we are in mod five, because every time you get to the number five, it resets and goes back to one. So if you want to learn more about this subject, look up modular arithmetic. It's kind of a fancy sounding name, but it's really not that difficult. It's just saying that every time you get past whatever number you're modding by, you go down to zero. So if for instance, you are working in mod 10, then you have zero, one, two, three, four, five, six, seven, eight, nine, and then it goes back to zero. And it's that sort of pattern. So modular arithmetic is a good thing to know if you're trying to get into abstract math in the future, stuff like number theory and abstract algebra, but that's way beyond the scope of this problem. For our intents and purposes, it's just helpful to think of this as 40,300 is equal to zero mod five, which means it's the same as making zero moves. So the square number is one. So we conclude that our final answer is A. Problem number 29, all the powers of three and all the numbers that can be written as a finite sum of different powers of three are arranged in the following increasing sequence. What will the hundredth element of the sequence be? So to start this off, before even thinking about the problem really, we can eliminate answer choice E because three to the hundredth will definitely be somewhere later down the line. If we were including only powers of three, then three to the hundredth would be the hundredth element. Well, actually it would be the hundred and first element because we also have three to the zeroth power, one here, but we're also including even more elements. So clearly E is not going to be the right choice. Now, how do we work with the other four choices? Well, since each of these can be expressed as a finite sum of some powers of three, we can go ahead and just list a whole bunch of powers of three that we know and see if we can write each of these answer choices as a sum of the powers. And in particular, if we go ahead and just write out a few of the powers, which are listed right here, and we see which of these can potentially add up to one of the answer choices, we'll see that most of these actually cannot be reached by adding up different powers of three. And in fact, you can go ahead and check each of these answer choices, A, C, and D, but you will not be able to find any sort of sum that works here, especially because, as the problem states, you're only allowed to use each of the powers once. Maybe it's not explicitly stated here, but it is implicitly indicated by the list that's given here because if, for instance, we were allowed to have some other number, if we were allowed to have multiple copies of a power, then it would actually just be the full list of numbers because we could just use three to the zeroth power over and over again, so one, two, three, four, five, six, and so on, and that would be a trivial problem. So each of the powers can only be used once. And once you check all of these three, for instance, if you look at 150 and you look at the different powers you have to work with, well, it's gonna be, we're working with three to the fourth, three cubed, three squared. It's kind of easy to see that they're not going to add up to 150. All you have to do is check. And you can just do that for all the answer choices. For answer choice B, we can go ahead and add up what looks like three to the sixth plus three to the fifth plus, let's see, it'll be 900 plus 60 plus 12, so that's 972, and then we could add three squared, which is nine, and that'll give us 981. So it is possible to come up with a way for the powers to add up to B. So that's gonna be our final answer. Problem number 30. Andrew, Michael, and Zach each roll a die. Andrew wins if he rolls one, two, or three. Michael wins if he rolls four or five, and Zach wins if he rolls a six. Andrew rolls first, then Michael, then Zach, and then Andrew rolls again, and so on. The game ends once one of them wins. What is the probability that Zach will win? So we can actually write this probability out as an infinite geometric series. In fact, here's what it'll look like, and let me explain what these fractions actually mean. So the first term in our series is 1 1 2 3 1 6. This represents the probability that Zach will be the one to win in round one. It's the probability that Andrew will lose, 1 1 2, times the probability Michael will lose, since Michael wins 1 3rd of the time, he will lose 2 3rds of the time, and then the probability that Zach will win. So this basically reflects the case where Zach wins in round one. What's the probability that everyone will lose in round one, and then Zach will lose in round two, or that Zach will win in round two? Well, that's the probability that everyone loses, 1 1 2 3rds times 5 6ths, times the probability that Zach wins in a round, which is 1 1 2 3rds times 1 6th. So basically what we're doing here is we're adding the probability that Zach wins on round one, on round two, on round three, on round four, all the way out to round infinity. And each time, we are multiplying once again by the probability that everybody loses in a round, because we want everyone, if we want Zach to win in the nth round, we need everyone to lose in the first n minus one rounds. So this is reflective of the first round's probability that Zach will win, this is the probability that Zach will win exactly in the second round, and this is the probability that Zach will win exactly in the third round. We just keep on adding a term that is multiplied by the probability that everyone will lose. But by definition, this is a geometric sequence where the term that we are multiplying each new term by is 1 1 2 3rds times 5 6ths. So looking up the definition of a geometric series, you have a starting term, and then you are adding that starting term over and over again, multiplied by powers of some constant term, which in this case, once again, is the probability that everyone loses. So identifying this as a geometric series where this multiplying term is 1 1 2 3rds times 5 6ths, we can use the formula that lets us calculate an infinite geometric series, and that's basically just a ratio. So it's going to be the probability that, it's going to be the starting term over one minus the multiplying term. That is the formula. The starting term over one minus the multiplying term. So our starting term is 1 1 2 3rds times 1 6th, which is equal to one over 18. And 1 1 2 3rds times 5 6ths is equal to five over 18. So one minus that is 13 over 18. And the ratio of these two, this fraction, the over 18s cancel out, so we just get one over 13. And again, this is a well-established formula that tells us what an infinite geometric series turns out to be. It needs to be a geometric series because a geometric series has both the multiplying term and the starting term. There are other formulas for different kinds of series, but this one in particular is good to know since geometric series come up a lot when you start studying higher levels of calculus. So this is sort of a good formula to keep in the back of your head if you wanna do a quick computation of an infinite series. And this gives us the final answer of one over 13. So our final answer is going to be D.

math problems

rhombus properties

graph analysis

function inverses

arithmetic progressions

probability games

integer constraints

geometric series