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Progressing to Arithmetic Progressions

BY KARISHMA, VERITAS PREP


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 \(a + (n-1)d\). The first term is given by \(a + (1-1)d = a\)

The second term is given by \(a + (2-1)d = a + d\) etc

The sum of n terms will be found by doing the following:

\(a + (a+d) + (a+2d) + … + (a+(n-1)d) = na + (d + 2d + 3d + … + (n-1)d)\) (summing the n first terms together and clubbing the common differences together in a bracket)

\(= na + d(1 + 2 + 3 + …(n-1))\) (taking d common)

\(= na + \frac{d(n-1)n}{2}\) (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.)

\(= \frac{n}{2} (2a + (n-1)d)\)

If you notice carefully, \(2a + (n-1)d\) can be re-written as: \(a + (a + (n-1)d)\) 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 \(t_n = t_{n-1} + 4\). 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 \(a + (n-1)d\)

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 \(t_1 = 4\), \(t_2 = 10\), …, \(t_n = t_{n-1} + 6\),…, 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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Time to Tackle Geometric Progressions!

BY KARISHMA, VERITAS PREP


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 non-zero 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 (M-1)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 \(t_{m-1} – t_{m} = 2*t_{m}\)

This tells us that \(t_m = \frac{t_{m-1}}{3}\)
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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How to Benefit from the GP Perspective

BY KARISHMA, VERITAS PREP


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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Re: NEW!!! Sequences Made Easy - All in One Topic! NEW!!! [#permalink]
[quote="Bunuel"]
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.

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 2n(2n+1)22n(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)=n2n(2n+1)–n(n+1)=n2.

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 \(n(2n+1) – n(n+1) = n^2\).

Hello!
I didn't get why do we need the sum of 2n to substract from and not only n. The sum of n consecutive terms minus the sum of n even consecutive terms will yield us the sum of odd terms. That's how i reasoned before reading explanation. However can't logically get the point why do we use the sum of 2n.

Could someone elaborate it please?
Thanks in advance!
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Re: NEW!!! Sequences Made Easy - All in One Topic! NEW!!! [#permalink]
LidiiaShchichko wrote:
Bunuel wrote:
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.

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 2n(2n+1)22n(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)=n2n(2n+1)–n(n+1)=n2.

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 \(n(2n+1) – n(n+1) = n^2\).

Hello!
I didn't get why do we need the sum of 2n to substract from and not only n. The sum of n consecutive terms minus the sum of n even consecutive terms will yield us the sum of odd terms. That's how i reasoned before reading explanation. However can't logically get the point why do we use the sum of 2n.

Could someone elaborate it please?
Thanks in advance!



If we take only N terms then we would recieve sum of N/2 odd terms.

So taking 2N terms gives us sum of N odd terms.
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