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NEW!!! Sequences Made Easy  All in One Topic! NEW!!!
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Updated on: 07 Apr 2018, 15:50
Sequences Made Easy  All in One Topic!Scrutinizing Sequences Here let’s dig into sequences on the GMAT. Let’s first understand what a sequence is (from Wikipedia): A sequence is an ordered list of objects. The number of terms it contains (possibly infinite) is called the length of the sequence. Unlike a set, order matters in a sequence, and exactly the same elements can appear multiple times at different positions in the sequence. Since order matters, (A, B, C) and (B, C, A) are two different sequences. (A series is the sum of the terms of a sequence but we will not deal with series today.) There are some special sequences e.g. arithmetic progressions and geometric progressions. We will deal with these in subsequent weeks. Today we will look at some generic sequence questions and will learn how to approach them. I will start with a very basic question. Mind you, most sequence questions will be higher level questions since sequence questions look complicated (even though they are very straight forward, believe me!). Let me show you using some questions from external sources: A note on notation: The first term of a sequence will be denoted by x(1), second term by x(2) and nth term by x(n). (If I want to show multiplication e.g. multiply x by 2, I will show it by writing x*2) Question 1: In a certain sequence, the term \(x_n\) is given by the formula \(x_n=2*x_{n1}\frac{1}{2}*x_{n2}\) for all \(n\geq{2}\). If \(x_0=3\) and \(x_1=2\), what is the value of \(x_3\)?(A) 2.5 (B) 3.125 (C) 4 (D) 5 (E) 6.75 Solution: This is a straight forward question event though the formula given is discomforting. Whenever you have a generic formula for the nth term of a sequence, plug in some numbers to see what pattern you get. \(x_0 = 3\) (given) \(x_1 = 2\) (given) If \(n = 2\), \(x_2 = 2*x_1 – \frac{1}{2}*x_0 = 2*2 – \frac{1}{2}*3 = \frac{5}{2}\) If \(n = 3\), \(x_3 = 2*x_2 – \frac{1}{2}*x_1 = 2*\frac{5}{2} – \frac{1}{2}*2 = 4\) Answer: C. This question is discussed HERE.I hope you agree that this was very simple. For \(x_3\), you needed \(x_2\). For \(x_2\), you needed \(x_1\) and \(x_0\), both of which you had! So it was a simple matter of quick substitution. Now let’s look a teeny bit complicated question Question 2: The infinite sequence \(a_1\), \(a_2\),… \(a_n\), … is such that \(a_1 = 4\), \(a_2 = 2\), \(a_3 = 6\), \(a_4 = 1\), and \(a_n = a_{n4}\) for \(n > 4\). If \(T = a_{10} + a_{11} + a_{12} + … a_{84} + a_{85}\), what is the value of T?(A) 119 (B) 120 (C) 121 (D) 126 (E) 133 Solution: We know the first four terms: \(a_1 = 4\), \(a_2 = 2\), \(a_3 = 6\), \(a_4 = 1\) Also it is given that \(a_n = a_{n4}\) i.e. the nth term is equal to the (n4)th term e.g. 5th term is equal to the 1st term. 6th term is equal to the 2nd term. 7th term is equal to the 3rd term etc. Hence, the sequence becomes: 4, 2, 6, 1, 4, 2, 6, 1, 4, 2, 6, 1 … (It is always helpful to write down the first few terms of the sequence. It helps you see the pattern.) The sequence has a cyclicity of 4 i.e. the terms repeat after every 4 terms (go back to the definition of sequence above – it says that the same element can appear multiple times at different positions). Therefore, first to fourth terms will form the first cycle, fifth to eighth terms will form the second cycle, ninth to twelfth terms will form the third cycle and so on… The sum of each group of 4 terms = 4 – 2 + 6 – 1 = 7 What will be the tenth term, \(a_{10}\)? A new cycle starts from \(a_9\) so \(a_9 = 4\). Then, \(a_{10}\) must be 2. \(a_{10} + a_{11} + a_{12}\) is the sum of last three terms of a cycle so this sum must be – 2 + 6 – 1 = 3 \(a_{13}\) to \(a_{16}\) is a complete cycle, \(a_{17}\) to \(a_{20}\) is another complete cycle and so on… The sum of each of the complete cycles is 7. How many such complete cycles will there be? The first complete cycle will end at \(a_{16}\), the second one at \(a_{20}\), the third one at \(a_{24}\) etc (i.e. at multiples of 4). The last complete cycle will end at a(84). How many complete cycles do we have here then? 16 = 4*4 and 84 = 4*21 so you start from the fourth multiple to the 21st multiple i.e. you have (21 – 4 + 1) = 18 total cycles. If you are confused about the ‘+1’ here, hang on – I will take it up at the end of this post. The sum of these 18 cycles will be 7*18 = 126 (I know the multiplication table of 18 as should you!) We still haven’t accounted for a(85), which will be the first term of the next cycle. The first term is 4. \(a_{10} + a_{11} + a_{12} + … a_{84} + a_{85} = 3 + 126 + 4 = 133 = T\) or you could just consider this: You have 18 complete cycles except for the first 3 terms and the last term of the sequence. The last term of the sequence is the first term of a cycle and the first three terms of the sequence are the last three terms of the cycle. So these four terms make one complete cycle. Therefore, instead of 18, you have 19 complete cycles. \(T = 7 * 19 = 133\) Answer (E). This question is discussed HERE.These were some basic sequences questions. I want to leave you with a sequence question from GMAT prep now. Try and work it out. We will look at its solution next week. Question 3: For every integer m from 1 to 10 inclusive, the \(m_{th}\) term of a certain sequence is given by \((1)^{(m+1)}*(\frac{1}{2})^m\). If T is the sum of the first 10 terms in the sequence, then T is:(A) greater than 2 (B) between 1 and 2 (C) between 0.5 and 1 (D) between 0.25 and 0.5 (E) less than 0.25 Note on the ‘+1’ above: How many numbers are there from 11 to 25, both inclusive? If your answer is 25 – 11 = 14, then you are wrong. When we say 25 – 11, we are saying that we have 25 numbers and we are throwing away 11 of them. But we want to keep the 11 (since we have both inclusive); we want all the numbers starting from 11 and ending at 25. We need to add 1 to the result to ensure that the 11 that we threw out, is retained. Consequently, the number of numbers from 11 to 25, both inclusive is 14 + 1 = 15.
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Originally posted by Bunuel on 02 Apr 2018, 00:11.
Last edited by Bunuel on 07 Apr 2018, 15:50, edited 9 times in total.
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Re: NEW!!! Sequences Made Easy  All in One Topic! NEW!!!
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02 Apr 2018, 00:28
Scrutinizing a 700+ Level Question on Sequences Here, I will take the question I gave you in the last post (and that is all we will tackle today!) It is a question from GMAT prep test so it is quite indicative of the tricky questions you might get on actual GMAT. Solving the question takes less than two minutes since the calculations required are negligible. However, if you start calculating the actual value, you could end up spending many painful minutes before giving up. I implore you to always remember that GMAT does not give you calculation intensive questions. Since they do not provide you with an HP12C, there will always be a logical solution – you will just need to think a little harder! Let’s get going then… Question 3: For every integer m from 1 to 10 inclusive, the \(m_{th}\) term of a certain sequence is given by \((1)^{(m+1)}*(\frac{1}{2})^m\). If T is the sum of the first 10 terms in the sequence, then T is:(A) greater than 2 (B) between 1 and 2 (C) between 0.5 and 1 (D) between 0.25 and 0.5 (E) less than 0.25 Solution:I agree that the expression for the mth term looks daunting. But as we discussed last week, the first step in a sequence question should be to write down the first few terms of the sequence. So let’s do that and see what we get. We get the first term by putting m = 1 First term = \((1)^{(1+1)}*(\frac{1}{2})^1 = \frac{1}{2}\) (Not so bad, eh?!) Second term = \((1)^{(2+1)}*(\frac{1}{2})^2 = \frac{1}{2}^2 = \frac{1}{4}\) Third term = \((1)^{(3+1)}*(\frac{1}{2})^3 = \frac{1}{2^3} = \frac{1}{8}\) Do we see a pattern? The tenth term will be \((1)^{(10+1)}*(\frac{1}{2})^10 = \frac{1}{2^10} = \frac{1}{1024}\) (You don’t need to calculate this of course. You can keep it as 2^10. I do suggest though that you should be good with the first 10 powers of 2, first 6 powers of 3, first 4 powers of 4 and first 3 powers of 5 to 10.) The sequence looks like this: 1/2, – 1/4, 1/8, – 1/16, … \(T = \frac{1}{2} – \frac{1}{4} + \frac{1}{8} – \frac{1}{16} +…+ \frac{1}{512} – \frac{1}{1024}\) Of course GMAT doesn’t expect us to calculate this. One could end up wasting a lot of precious time if one did. The trick is to know that we need to figure out the answer using some shrewdness. Fortunately (or unfortunately), GMAT software rewards cunning and craft! The problem is that we have positive and negative terms so it is very hard to say what the value of T will be. But the terms are not random. There is a positive term followed by a negative term which is then followed by a positive term and so on. Also, every subsequent term is smaller than the previous term (in fact it is a Geometric Progression but we don’t need to know that to solve this question. Nevertheless, we will take up GP too in a couple of weeks). We need to create some uniformity so that we can deduce something about T. We have 10 terms. If we couple them up, two terms each, we get 5 groups: \(T = (\frac{1}{2} – \frac{1}{4}) + (\frac{1}{8} – \frac{1}{16}) + … + (\frac{1}{512} – \frac{1}{1024})\) Tell me, can we say that each group is positive? From a larger number, you are subtracting a smaller number in each bracket. The first number is greater than the second number in each group e.g. 1/2 is greater than 1/4, therefore, (1/2 – 1/4) = 1/4 i.e. a positive number Similarly, (1/8 – 1/16) = 1/16, again a positive number. This means \(T = \frac{1}{4} + \frac{1}{16} +…. (all \ positives)\) Definitely this sum, T, is greater than 1/4 i.e. 0.25 So we can rule out option (E). But we still have to choose one out of the four remaining options. Now, let’s group the terms in another way. \(T = \frac{1}{2} + (\frac{1}{4} + \frac{1}{8}) + (\frac{1}{16} + \frac{1}{32}) … – \frac{1}{1024}\) You leave out the first term and start grouping two terms at a time. The last term will be left alone too! You will be able to make four groups with the 8 terms in the middle. Now look closely at each group: The first term is a negative number with a higher absolute value while the second term is a smaller positive number so the sum will give you a negative number, e.g.: \((\frac{1}{4} + \frac{1}{8}) = \frac{1}{8}\) \((\frac{1}{16} + \frac{1}{32}) = \frac{1}{32}\) etc This means \(T = \frac{1}{2} – \frac{1}{8} – \frac{1}{32} … \frac{1}{1024}\) All 4 of the groups will give you a negative number and the last term is also negative. Since the first term is 1/2 i.e. 0.5, we can say that the sum T will be less than 0.5 since all the other terms are negative. So the sum, T, must be more than 0.25 but less than 0.5. Answer has to be option (D). This question is discussed HERE.There are other ways of arriving at the answer here. We will look at it from the Geometric Progression perspective some time later. I will leave you now with a question I saw somewhere once. Let’s see if you can use your craft to arrive at the answer in a minute! (absolutely doable) Question: In the infinite sequence A, \(A_n = x^{(n1)} + x^n + x^{(n+1)} + x^{(n+2)} + x^{(n+3)}\) where x is a positive integer constant. For what value of n is the ratio of \(A_n\) to \(x(1+x(1+x(1+x(1+x))))\) equal to \(x^5\)?(A) 8 (B) 7 (C) 6 (D) 5 (E) 4 This question is discussed HERE.
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Re: NEW!!! Sequences Made Easy  All in One Topic! NEW!!!
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02 Apr 2018, 00:57
Progressing to Arithmetic Progressions Today’s topic is Arithmetic Progression (AP). An AP is a sequence of numbers such that the difference between the consecutive terms is constant. For example: 2, 4, 6, 8… 1, 10, 21, 32… 4, 3, 2, 1, 0, 1, 2, 3 1/2, 3/2, 5/2, 7/2… and so on. Note that the numbers could be increasing or decreasing. As long as the difference between the consecutive terms is constant, it is an AP. The general form of an AP is given by \(a\), \(a+d\), \(a+2d\), \(a+3d\)… I hope you see how we get it. \(a\) is the first term and \(d\) is the common difference. The nth term has the form \(a + (n1)d\). The first term is given by \(a + (11)d = a\) The second term is given by \(a + (21)d = a + d\) etc The sum of n terms will be found by doing the following: \(a + (a+d) + (a+2d) + … + (a+(n1)d) = na + (d + 2d + 3d + … + (n1)d)\) (summing the n first terms together and clubbing the common differences together in a bracket) \(= na + d(1 + 2 + 3 + …(n1))\) (taking d common) \(= na + \frac{d(n1)n}{2}\) (this is because the sum of n consecutive integers starting from 1 is given by n(n+1)/2. We will work on this concept in detail next week.) \(= \frac{n}{2} (2a + (n1)d)\) If you notice carefully, \(2a + (n1)d\) can be rewritten as: \(a + (a + (n1)d)\) i.e. the sum of first and last terms. So basically the sum of n terms of the AP is \(n * \frac{(First \ term + Last \ term)}{2}\) which is the same as \(number \ of \ terms * Average \ of \ the \ first \ and \ the \ last \ terms\).Questions on APs are very simple. Sometimes, people just don’t realize that the question is based on an AP. The questions in GMAT may not give you that the sequence is an AP but it is not tough to figure out. Let us look at an example now. Question 1: The \(n_{th}\) term, \(t_n\), of a certain sequence is defined as \(t_n = t_{n1} + 4\). If \(t_1 = 11\), then \(t_{82} =\)(A) 313 (B) 317 (C) 320 (D) 321 (E) 340 Solution: The given relation says that every nth term is 4 more than the previous term. So basically, it tells us that the sequence is an AP. Whew! (An AP is very easy to work with) What is the nth term of an AP? It’s \(a + (n1)d\) What is the 82nd term of this AP? It’s \(11 + 81*4 = 313\) Answer (A). This question is discussed HERE.Question 2: If S is the infinite sequence such that \(t_1 = 4\), \(t_2 = 10\), …, \(t_n = t_{n1} + 6\),…, what is the sum of all the terms from \(t_{10}\) to \(t_{18}\)?(A) 671 (B) 711 (C) 738 (D) 826 (E) 991 Solution: Again, since the nth term is 6 more than the previous term, it is an AP. We need to find the following sum: \(t_{10} + t_{11} + t_{12} + … + t_{18}\) But we only know how to find the sum starting from t(1). Let’s manipulate what we have to find a little to make it similar to what we know. \(t_{10} + t_{11} + t_{12} + … + t_{18} = Sum \ of \ first \ 18 \ terms – Sum \ of \ first \ 9 \ terms\) Sum of first 18 terms = \((\frac{18}{2})(2*4 + 17*6)\) Sum of first 9 terms = \((\frac{9}{2})(2*4 + 8*6)\) \(t_{10} + t_{11} + t_{12} + … + t_{18} = (\frac{18}{2})(2*4 + 17*6) – (\frac{9}{2})(2*4 + 8*6)\) \(= 990 – 252 = 738\) Answer (C). This question is discussed HERE.As you see, AP questions are easy to work with. Next week, we will discuss some properties of a specific type of AP i.e. consecutive integers. Till then, keep practicing!
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Re: NEW!!! Sequences Made Easy  All in One Topic! NEW!!!
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02 Apr 2018, 01:08
Special Arithmetic Progressions Above, we looked at some basic formulas related to Arithmetic Progressions. Here, we will look at a particular (and related) type of Arithmetic Progression — Consecutive Integers. Look at the following three sequences: S1 = 3, 4, 5, 6, 7 S2 = 1, 0, 1, 2, 3, 4, 5, 6, 7 S3 = 1, 2, 3, 4, 5, 6, 7, 8, 9 All of them are APs of consecutive integers so every formula we looked at last week is applicable here. Sum of an AP \(= (\frac{n}{2})(2a + (n1)d)\) Sum of the terms of S1 \(= (\frac{5}{2})(2*3 + (51)1) = 25\) Sum of the terms of S2 \(= (\frac{9}{2})(2*(1) + (91)1) = 27\) (or treat it as a sum of 2, 3, 4, 5, 6, 7) Sum of the terms of S3 \(= (\frac{9}{2})(2*1 + (91)1) = 45\) We will pay special attention to S3 i.e. an AP of consecutive integers starting from 1 Sum of the terms in this case \(= (\frac{n}{2})(2*1 + (n1)*1)\) (since a = 1 and d = 1) \((\frac{n}{2})(2*1 + (n1)*1) = (\frac{n}{2})(n+1) = \frac{n*(n+1)}{2}\) This is a formula we should be very comfortable with: sum of first n terms of a sequence of consecutive integers starting from 1 \(= \frac{n*(n+1)}{2}\). It is very useful in a lot of situations. We can also derive a lot of other relations using this single formula. For Example:Example 1. What is the sum of positive consecutive even integers starting from 2?– It is n(n+1) where n is the number of even integers (remember, n is not the last term; it is the number of total even integers). – How? – Say there are n numbers in the sequence and we want to find their sum: 2 + 4 + 6 + 8 + … We take 2 common out of them to get 2(1 + 2 + 3+ 4…). This is twice the sum of n consecutive integers starting from 1. So Sum \(= 2*\frac{n(n+1)}{2}= n(n+1)\). Example 2. What is the sum of positive consecutive odd integers starting from 1?– It is n^2 where n is the number of odd integers (again, n is not the last term; it is the number of total odd integers). – How? – Say there are n numbers in the sequence and we want to find their sum: 1 + 3 + 5 + 7 + … Sum of 2n consecutive integers (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8… ) will be \(\frac{2n(2n+1)}{2}\) and sum of n even consecutive integers (2 + 4 + 6 + 8 + …) starting from 2 will be n(n+1) so sum of the leftover n consecutive odd integers will be \(n(2n+1) – n(n+1) = n^2\). With this background, let’s look at a question similar to an OG12 question: Question 1: What is the sum of all the even integers between 99 and 401 ?(A) 10,100 (B) 20,200 (C) 37,750 (D) 40,200 (E) 45,150 Solution: We can solve this question in multiple ways. Required Sum = 100 + 102 + 104 + 106 + … + 400 Method 1:We know the sum of consecutive even integers but only when they start from 2. So what do we do? We find the sum of even integers starting from 2 till 400 and subtract the sum of even integers starting from 2 till 98 from it! Note that we subtract even numbers till 98 because 100 is a part of our series. How many even integers are there from 2 to 400? I hope you agree that we will have 200 even integers in this range (both inclusive) Sum of these 200 integers = 200(201) How many even integers are there from 2 to 98? Now we have 49 even integers here. Sum of these 49 even integers = 49(50) What is the sum of integers from 100 to 400? It will be 200(201) – 49(50) = 40200 – 2450 = 37750 Method 2:100 + 102 + 104 + 106 + … + 400 = 2( 50 + 51 + 52 + 53 + … + 200) (We take out 2 common and find the sum in brackets) Sum in brackets = 50 + 51 + 52 + 53 + … + 200 We know the sum of consecutive integers but only when they start from 1. So we find the sum of first 200 numbers and subtract the sum of first 49 numbers from it. That will give us the sum of numbers from 50 to 200. Note that we subtract 49 numbers because 50 is a part of our series. Sum of 1 + 2 + 3 + … + 200 = 200*201/2 = 20100 Sum of 1 + 2 + 3 + … + 49 = 49*50/2 = 1225 (I am doing these calculations here only for clarity. Normally, I would like to carry all these till the last step, then take common, divide by whatever I can etc so that I have very few actual calculations left.) Therefore, 50 + 51 + 52 + 53 + … + 200 = 20100 – 1225 = 18875 Then 100 + 102 + 104 + 106 + … + 400 = 2*18875 = 37750 Answer (C). This question is discussed HERE.I hope you see that questions on arithmetic progressions are generally quite simple. Next week, we will move on to geometric progressions. Till then, keep practicing!
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Re: NEW!!! Sequences Made Easy  All in One Topic! NEW!!!
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02 Apr 2018, 01:26
Time to Tackle Geometric Progressions! Let’s look at geometric progressions (GP) now. Before I start, let me point out that GMAT is unlikely to give you a statement which looks like this: “If S denotes a geometric progression whose first term is… ” GMAT will not test your knowledge of GP (i.e. you don’t really need to learn the formulas of the sum of n terms of a GP or sum of infinite terms of a GP etc) though it may give you a sequence which is a geometric progression and ask you questions on it. You will be able to solve the question without using the formulas but recognizing a GP can help you deal with such questions in an efficient manner. That is the reason we are discussing GPs today. For those of you who are wondering what exactly a GP is, let me begin by giving you the definition. (I will quote Wikipedia here.) A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed nonzero number called the common ratio. For example, the sequence 2, 6, 18, 54, … is a geometric progression with common ratio 3 (each term after the first is obtained by multiplying the previous term by 3). Similarly 10, 5, 5/2, 5/4, … is a geometric sequence with common ratio 1/2.The first term of a GP is generally denoted by a and the common ratio is denoted by r. So the general form of a GP is: \(a\), \(ar\), \(ar^2\), \(ar^3\), …, \(ar^{(n – 1)}\) (The GP has n terms here.) Sum of n terms of a GP is given by \(\frac{a*(1 – r^n)}{(1 – r)}\). Sum of all the terms of an infinite GP is given by \(\frac{a}{(1 – r)}\), for r < 1. We generally look at the derivation of the formula to help us understand it better (and hence remember it) but the derivation of this formula is more mathematical and less intuitive so we will not discuss it here. If you are interested in the derivation, check out the Wikipedia link given above. We will directly jump to GMAT relevant questions now and see how they can be solved without the aid of the formulas and how they can be solved with the formulas. First we will look at a simple DS question which deals with GP but doesn’t mention it in the question stem. The challenge is to figure out that it is a GP. Once you do, you can solve it in a few moments. Question 1: In a certain sequence, when you subtract the Mth element from the (M1)th element, you get twice the Mth element (M is any positive integer). What is the fourth element of this sequence?
(1) The first element of the sequence is 1. (2) The third element of the sequence is 1/9.Solution: The question stem doesn’t tell us that the sequence is a geometric progression. It only tells us that \(t_{m1} – t_{m} = 2*t_{m}\) This tells us that \(t_m = \frac{t_{m1}}{3}\) Every subsequent term should be a third of the previous term. This means that the sequence is a GP with common ratio = 1/3. So the GP looks something like this: \(a\), \(\frac{a}{3}\), \(\frac{a}{9}\), \(\frac{a}{27}\) … Statement 1: The first term of the sequence is 1. Now we know all the terms of the sequence. 1, 1/3, 1/9, 1/27… The fourth element is 1/27 so this statement alone is sufficient. Statement 2: The third term of the sequence is 1/9. The third term of our GP is a/9. If a/9 = 1/9, this implies that a = 1. Hence the fourth term = 1/27. This statement alone is also sufficient to answer the question. Answer (D). This question is discussed HERE.Now we will look at a trickier question. Question 2: What is the value of \(7 + 6*7 + 6*7^2 + 6*7^3 + 6*7^4 + 6*7^5 + 6*7^6\)?(A) 6^7 (B) 6^9 (C) 7^7 (D) 7^8 (E) 7^9 Solution: First let’s solve the question without using the GP formula. Method 1:\(S = 7 + 6*7 + 6*7^2 + 6*7^3 + 6*7^4 + 6*7^5 + 6*7^6\) \(S = 7*(1 + 6) + 6*7^2 + 6*7^3 + 6*7^4 + 6*7^5 + 6*7^6\) (Take 7 common from the first two terms) \(S = 7^2 + 6*7^2 + 6*7^3 + 6*7^4 + 6*7^5 + 6*7^6\) \(S = 7^2 * (1 + 6) + 6*7^3 + 6*7^4 + 6*7^5 + 6*7^6\) (Take 7^2 common from the first two terms) \(S = 7^3 + 6*7^3 + 6*7^4 + 6*7^5 + 6*7^6\) I hope you see where we are going with this. The last step would be: \(S = 7^6 * (1 + 6) = 7^7\) Answer (C). This question is discussed HERE.Method 2:Except for the first term of the sequence, the rest of the sequence is a GP with first term as 6*7 and the common ratio as 7. \(S = 7 + 6*7 + 6*7^2 + 6*7^3 + 6*7^4 + 6*7^5 + 6*7^6 = 7 + GP\) There are 6 terms in the GP. Sum of the \(GP = \frac{a*(1 – r^n)}{(1 – r)}= \frac{6*7*(1 – 7^6)}{(1 – 7)} = \frac{6*7 * (7^6 – 1)}{6} = 7^7 – 7\) Substituting this sum back in S, we get \(S = 7 + 7^7 – 7 = 7^7\) Answer (C)The first method is not difficult. It is just hard to figure out when you are under time pressure. A GP is easy to notice and you know exactly how to handle it. I will not advice you to use one method over the other – both are equally valid and good so use whatever works for you. The only thing is that knowing how to deal with GPs can help save time. It may not be very apparent in this question but I will leave you with a question in which it will be apparent! It is something similar to the 700+ level GMAT prep test question we saw while working on arithmetic progressions. We will discuss this question in detail next week. Question 3: For every integer n from 1 to 200, inclusive, the nth term of a certain sequence is given by \((1)^n*2^{(n)}\). If N is the sum of the first 200 terms in the sequence, then N is(A) less than 1 (B) between 1 and 1/2 (C) between 1/2 and 0 (D) between 0 and 1/2 (E) greater than 1/2
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Re: NEW!!! Sequences Made Easy  All in One Topic! NEW!!!
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02 Apr 2018, 01:47
How to Benefit from the GP Perspective Last week, at the end of the Geometric Progression (GP) post, I gave you a question to figure out. I hope some of you did try it. Today we will discuss the question in detail and look at two different approaches – one without using GP formula (we discussed this approach in a previous post) and another with the formula. As I said before, you can solve every sequence question on GMAT without using the formulas we are discussing. We are still investing time in these formulas so that we can save some in the actual exam. Let me show you how. Question 3: For every integer n from 1 to 200, inclusive, the nth term of a certain sequence is given by \((1)^n*2^{(n)}\). If N is the sum of the first 200 terms in the sequence, then N is:(A) less than 1 (B) between 1 and 1/2 (C) between 1/2 and 0 (D) between 0 and 1/2 (E) greater than 1/2 Solution: The question seems a little intimidating since the options give you ranges. This means that it is probably hard to find the exact value and that’s just not good. Anyway, let’s begin by doing what we know we should do with every sequence question if possible: we should write out the first few terms of the sequence. First term: \((1)^1*2^{(1)} = \frac{1}{2}\) Second term: \((1)^2*2^{(2)} = \frac{1}{4}\) Third term: \((1)^3*2^{(3)} = \frac{1}{8}\) and so on… The sequence looks like this: \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \(\frac{1}{16}\), \(\frac{1}{32}\)… \(N = \frac{1}{2} + \frac{1}{4} – \frac{1}{8} + \frac{1}{16} – \frac{1}{32} …\) (200 terms) Method 1:Say, we do not want to work with GPs. Let’s sum the pair of consecutive terms. (Guess you are reminded of the 700+ level GMAT prep question we discussed a couple of weeks back.) \(N = (\frac{1}{2} + \frac{1}{4}) + ( \frac{1}{8} + \frac{1}{16}) + ( \frac{1}{32} + \frac{1}{64}) + …\) \(\frac{1}{2} + \frac{1}{4} = \frac{1}{4}\) \(\frac{1}{8} + \frac{1}{16} = \frac{1}{16}\) \(\frac{1}{32} + \frac{1}{64} = \frac{1}{64}\) … \(N = \frac{1}{4} – \frac{1}{16} – \frac{1}{64} …\) We can see that the sum of all these terms will be less than 1/4 since all the rest of the terms are negative too. \(N < \frac{1}{4}\) Let’s look at it in another way. Let’s leave the first term and sum the pair of consecutive terms thereafter. \(N = \frac{1}{2} + (\frac{1}{4} – \frac{1}{8}) + (\frac{1}{16} – \frac{1}{32}) + …\) \(N = \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + …\) Since all the terms after the first one are positive, N must be greater than 1/2. Therefore, \(\frac{1}{2} < N < \frac{1}{4}\). Hence, of the given ranges, N must lie between 1/2 and 0. Answer (C). This question is discussed HERE.Let’s look at a more straight forward method now. Method 2:Let’s say we recognize that the given sequence is a GP. \(N = \frac{1}{2} + \frac{1}{4} – \frac{1}{8} + \frac{1}{16} – \frac{1}{32} …\) (200 terms) The first term is 1/2 and the common ratio is 1/2. Sum of 200 terms of a \(GP = \frac{a(1 – r^n)}{(1 – r)} = \frac{(\frac{1}{2})(1 – (\frac{1}{2})^{200})}{1 – (\frac{1}{2})}\) Notice that in the bracket \((1 – (\frac{1}{2})^{200})\), \((\frac{1}{2})^{200}\) is a very very small number compared to 1 so the value of the bracket is approximately 1. Sum of the GP \(= \frac{(\frac{1}{2})(1)}{1 (\frac{1}{2})} = \frac{1}{3}\) Since 1/3 lies between 1/2 and 0, answer is (C). The second method was much faster and much more mechanical than the first one. I don’t particularly encourage mechanical thought process but during the exam thinking up innovative methods is a little hard if you haven’t trained your mind to do so. Hence, knowing how to deal with APs and GPs is a good idea.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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