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NEW!!! Sequences Made Easy - All in One Topic! NEW!!!
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Updated on: 07 Apr 2018, 16:50
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Sequences Made Easy - All in One Topic!
CONTENT
Scrutinizing Sequences
BY KARISHMA, VERITAS PREP
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 \(x_n=2*x_{n-1}-\frac{1}{2}*x_{n-2}\) 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 \(a_n = a_{n-4}\) 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 \(a_n = a_{n-4}\) i.e. the nth term is equal to the (n-4)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.
Re: NEW!!! Sequences Made Easy - All in One Topic! NEW!!!
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02 Apr 2018, 01:28
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Scrutinizing a 700+ Level Question on Sequences
BY KARISHMA, VERITAS PREP
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 do-able)
Question: In the infinite sequence A, \(A_n = x^{(n-1)} + x^n + x^{(n+1)} + x^{(n+2)} + x^{(n+3)}\) 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.
Re: NEW!!! Sequences Made Easy - All in One Topic! NEW!!!
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02 Apr 2018, 02:08
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Special Arithmetic Progressions
BY KARISHMA, VERITAS PREP
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 \(= (\frac{n}{2})(2a + (n-1)d)\)
Sum of the terms of S1 \(= (\frac{5}{2})(2*3 + (5-1)1) = 25\)
Sum of the terms of S2 \(= (\frac{9}{2})(2*(-1) + (9-1)1) = 27\) (or treat it as a sum of 2, 3, 4, 5, 6, 7)
Sum of the terms of S3 \(= (\frac{9}{2})(2*1 + (9-1)1) = 45\)
We will pay special attention to S3 i.e. an AP of consecutive integers starting from 1
Sum of the terms in this case \(= (\frac{n}{2})(2*1 + (n-1)*1)\) (since a = 1 and d = 1)
\((\frac{n}{2})(2*1 + (n-1)*1) = (\frac{n}{2})(n+1) = \frac{n*(n+1)}{2}\)
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!