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# Math : Sequences & Progressions

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Math : Sequences & Progressions  [#permalink]

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Updated on: 10 Dec 2017, 07:49
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Sequences & Progressions

This post is a part of [GMAT MATH BOOK]

created by: shrouded1

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Definition

Sequence : It is an ordered list of objects. It can be finite or infinite. The elements may repeat themselves more than once in the sequence, and their ordering is important unlike a set

Arithmetic Progressions

Definition
It is a special type of sequence in which the difference between successive terms is constant.

General Term
$$a_n = a_{n-1} + d = a_1 + (n-1)d$$
$$a_i$$ is the ith term
$$d$$ is the common difference
$$a_1$$ is the first term

Defining Properties
Each of the following is necessary & sufficient for a sequence to be an AP :
• $$a_i - a_{i-1} =$$ Constant
• If you pick any 3 consecutive terms, the middle one is the mean of the other two
• For all i,j > k >= 1 : $$\frac{a_i - a_k}{i-k} = \frac{a_j-a_k}{j-k}$$

Summation
The sum of an infinite AP can never be finite except if $$a_1=0$$ & $$d=0$$
The general sum of a n term AP with common difference d is given by $$\frac{n}{2}(2a+(n-1)d)$$
The sum formula may be re-written as $$n * Avg(a_1,a_n) = \frac{n}{2} * (FirstTerm+LastTerm)$$

Examples
1. All odd positive integers : {1,3,5,7,...} $$a_1=1, d=2$$
2. All positive multiples of 23 : {23,46,69,92,...} $$a_1=23, d=23$$
3. All negative reals with decimal part 0.1 : {-0.1,-1.1,-2.1,-3.1,...} $$a_1=-0.1, d=-1$$

Geometric Progressions

Definition
It is a special type of sequence in which the ratio of consequetive terms is constant

General Term
$$b_n = b_{n-1} * r = a_1 * r^{n-1}$$
$$b_i$$ is the ith term
$$r$$ is the common ratio
$$b_1$$ is the first term

Defining Properties
Each of the following is necessary & sufficient for a sequence to be an GP :
• $$\frac{b_i}{b_{i-1}} =$$ Constant
• If you pick any 3 consecutive terms, the middle one is the geometric mean of the other two
• For all i,j > k >= 1 : $$(\frac{b_i}{b_k})^{j-k} = (\frac{b_j}{b_k})^{i-k}$$

Summation
The sum of an infinite GP will be finite if absolute value of r < 1
The general sum of a n term GP with common ratio r is given by $$b_1*\frac{r^n - 1}{r-1}$$
If an infinite GP is summable (|r|<1) then the sum is $$\frac{b_1}{1-r}$$

Examples
1. All positive powers of 2 : {1,2,4,8,...} $$b_1=1, r=2$$
2. All negative powers of 4 : {1/4,1/16,1/64,1/256,...} $$b_1=1/4, r=1/4, sum=\frac{1/4}{(1-1/4)}=(1/3)$$

Harmonic Progressions

Definition
It is a special type of sequence in which if you take the inverse of every term, this new sequence forms an HP

Important Properties
Of any three consecutive terms of a HP, the middle one is always the harmonic mean of the other two, where the harmonic mean (HM) is defined as :
$$\frac{1}{2} * (\frac{1}{a} + \frac{1}{b}) = \frac{1}{HM(a,b)}$$
Or in other words :
$$HM(a,b) = \frac{2ab}{a+b}$$

Each progression provides us a definition of "mean" :

Arithmetic Mean : $$\frac{a+b}{2}$$ OR $$\frac{a1+..+an}{n}$$
Geometric Mean : $$\sqrt{ab}$$ OR $$(a1 *..* an)^{\frac{1}{n}}$$
Harmonic Mean : $$\frac{2ab}{a+b}$$ OR $$\frac{n}{\frac{1}{a1}+..+\frac{1}{an}}$$

For all non-negative real numbers : AM >= GM >= HM

In particular for 2 numbers : AM * HM = GM * GM

Example :
Let a=50 and b=2,
then the AM = (50+2)*0.5 = 26 ;
the GM = sqrt(50*2) = 10 ;
the HM = (2*50*2)/(52) = 3.85
AM > GM > HM
AM*HM = 100 = GM^2

Misc Notes
A subsequence (any set of consequutive terms) of an AP is an AP

A subsequence (any set of consequutive terms) of a GP is a GP

A subsequence (any set of consequutive terms) of a HP is a HP

If given an AP, and I pick out a subsequence from that AP, consisting of the terms $$a_{i1},a_{i2},a_{i3},...$$ such that $$i1,i2,i3$$ are in AP then the new subsequence will also be an AP

For Example : Consider the AP with $$a_1=1, d=2$$ {1,3,5,7,9,11,...}, so a_n=1+2*(n-1)=2n-1
Pick out the subsequence of terms $$a_5,a_{10},a_{15},...$$
New sequence is {9,19,29,...} which is an AP with $$a_1=9$$ and $$d=10$$

If given a GP, and I pick out a subsequence from that GP, consisting of the terms $$b_{i1},b_{i2},b_{i3},...$$ such that $$i1,i2,i3$$ are in AP then the new subsequence will also be a GP

For Example : Consider the GP with $$b_1=1, r=2$$ {1,2,4,8,16,32,...}, so b_n=2^(n-1)
Pick out the subsequence of terms $$b_2,b_4,b_6,...$$
New sequence is {4,16,64,...} which is a GP with $$b_1=4$$ and $$r=4$$

The special sequence in which each term is the sum of previous two terms is known as the fibonacci sequence. It is neither an AP nor a GP. The first two terms are 1. {1,1,2,3,5,8,13,...}

In a finite AP, the mean of all the terms is equal to the mean of the middle two terms if n is even and the middle term if n is odd. In either case this is also equal to the mean of the first and last terms

Some examples

Example 1
A coin is tossed repeatedly till the result is a tails, what is the probability that the total number of tosses is less than or equal to 5 ?

Solution
P(<=5 tosses) = P(1 toss)+...+P(5 tosses) = P(T)+P(HT)+P(HHT)+P(HHHT)+P(HHHHT)
We know that P(H)=P(T)=0.5
So Probability = 0.5 + 0.5^2 + ... + 0.5^5
This is just a finite GP, with first term = 0.5, n=5 and ratio = 0.5. Hence :
Probability = $$0.5 * \frac{1-0.5^5}{1-0.5} = \frac{1}{2} * \frac{\frac{31}{32}}{\frac{1}{2}} = \frac{31}{32}$$

Example 2
In an arithmetic progression a1,a2,...,a22,a23, the common difference is non-zero, how many terms are greater than 24 ?
(1) a1 = 8
(2) a12 = 24

Solution
(1) a1=8, does not tell us anything about the common difference, so impossible to say how many terms are greater than 24
(2) a12=24, and we know common difference is non-zero. So either all the terms below a12 are greater than 24 and the terms above it less than 24 or the other way around. In either case, there are exactly 11 terms either side of a12. Sufficient

Example 3
For positive integers a,b (a<b) arrange in ascending order the quantities a, b, sqrt(ab), avg(a,b), 2ab/(a+b)

Solution
Using the inequality AM>=GM>=HM, the solution is :
a <= 2ab/(a+b) <= Sqrt(ab) <= Avg(a,b) <= b

Example 4
For every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) *(1/2^k). 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 1/2 and 1
d)between 1/4 and 1/2
e)less than 1/4.

Solution
The sequence given has first term 1/2 and each subsequent term can be obtained by multiplying with -1/2. So it is a GP. We can use the GP summation formula
$$S=b\frac{1-r^n}{1-r}=\frac{1}{2} * \frac{1-(-1/2)^{10}}{1-(-1/2)} = \frac{1}{3} * \frac{1023}{1024}$$
1023/1024 is very close to 1, so this sum is very close to 1/3

Example 5
The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression?
A. 300
B. 120
C. 150
D. 170
E. 270

Solution
$$a_4+a_12=20$$
$$a_4=a_1+3d, a_12=a_1+11d$$
$$2a_1+14d=20$$
Now we need the sum of first 15 terms, which is given by :
$$\frac{15}{2} (2a_1 + (15-1)d) = \frac{15}{2} * (2a_1+14d) = 150$$

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Originally posted by shrouded1 on 28 Sep 2010, 16:32.
Last edited by Bunuel on 10 Dec 2017, 07:49, edited 12 times in total.
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Re: Math : Sequences & Progressions  [#permalink]

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28 Sep 2010, 17:25
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Great initiative.

Suggestions :

1. Include AM, GM , HM included between 2 numbers.
2. use a_{n} with math tag to get $$a_{n}$$
3. Include more examples.

This post,along with the algebra, is a good initiative to fill the important topics of Math Book that were not included earlier.
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Re: Math : Sequences & Progressions  [#permalink]

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28 Sep 2010, 22:33
Thanks for putting it here!
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Re: Math : Sequences & Progressions  [#permalink]

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29 Sep 2010, 10:27
2
gurpreetsingh wrote:
Great initiative.

Suggestions :

1. Include AM, GM , HM included between 2 numbers.
2. use a_{n} with math tag to get $$a_{n}$$
3. Include more examples.

This post,along with the algebra, is a good initiative to fill the important topics of Math Book that were not included earlier.

Check
Check
Check

Let me know what else ?
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Re: Math : Sequences & Progressions  [#permalink]

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29 Sep 2010, 19:30
Great initiative. +1 to you.
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30 Sep 2010, 05:25
Great one
Kudos to you !
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Re: Math : Sequences & Progressions  [#permalink]

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07 Oct 2010, 01:33
1
added a couple more solved GMAT style questions to the end
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07 Oct 2010, 01:56
Good to know. It can definitely same valuable time.
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Re: Math : Sequences & Progressions  [#permalink]

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18 Oct 2010, 00:10
1
This is great post but I was just wondering whether we need to know these concepts for GMAT
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18 Oct 2010, 23:49
There are several questions in the OG that use these concepts. So I think its good to know all this. Plus if you search through the forums you'll find several Qs on sequences and progressions as well
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Re: Math : Sequences & Progressions  [#permalink]

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29 Oct 2010, 21:19
Valuable resource.
Kudos +1
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Re: Math : Sequences & Progressions  [#permalink]

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02 Nov 2010, 11:53
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I had a question. I am not sure if this works: "In case of n numbers : AM * HM = GM^n".

This seems to work for 2 numbers but for more than 2, it seems to break, please let me know if I am missing something.
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02 Nov 2010, 12:28
nitantsharma wrote:
I had a question. I am not sure if this works: "In case of n numbers : AM * HM = GM^n".

This seems to work for 2 numbers but for more than 2, it seems to break, please let me know if I am missing something.

You are correct, this should only hold for special case n=2. Thanks for pointing out
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Re: Math : Sequences & Progressions  [#permalink]

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11 Nov 2010, 21:52
How can we get the formular for sumation of GEOMETRIC PROGRESSION. Please, prove, so that I do not have to remember the formular but to know the way to get the formular and so can solve the relative questions.
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12 Nov 2010, 02:34
 ! This proof is beyond the scope of the GMAT

The proof below is based on mathematical induction

To prove : The sum of an n term GP : $$b,br,br^2,...,br^{n-1}$$ is $$b*\frac{r^n-1}{r-1}$$

P(1 term) : The sum of the GP {b} is $$b*\frac{r^1-1}{r-1}=b$$. Which is true trivially

P(n terms) : Let the sum of an n term GP : $$b,br,...,br^{n-1}$$ be $$b*\frac{r^n-1}{r-1}$$

P(n+1 terms) : Consider the n+1 term GP : $$b,br,....,br^n$$
Sum of this GP = Sum of n term GP + $$br^n$$ = $$b*\frac{r^n-1}{r-1} + br^n$$
Sum = $$\frac{b}{r-1} * (r^n - 1 + r^n(r-1))$$
=$$\frac{b}{r-1} *(r^{n+1}-1)$$

Hence P(1) is true
And if we assume P(n) true P(n+1) is true
By mathematical induction P(k) must be true for all k>=1
Hence, proved
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Re: Math : Sequences & Progressions  [#permalink]

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17 Jan 2011, 03:09

I think the formula for calculating the sum of n consecutive numbers should be:

$$(lastterm - firstterm)*(lastterm - firstterm + 1)/2$$
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24 Jan 2011, 10:48
nonameee wrote:

I think the formula for calculating the sum of n consecutive numbers should be:

$$(lastterm - firstterm)*(lastterm - firstterm + 1)/2$$

1,2,3,4,5

let's apply your formula on the above series.

$$\frac{(5-1)*(5-1+1)}{2} = \frac{4*5}{2} = 10$$

this is not correct. let's try another formula.

$$\frac{n}{2}(firstterm + lastterm)$$

$$\frac{5}{2}(5+1) = 15$$

$$\frac{1}{2}(firstterm + lastterm)$$ basically gives you the avg of the series. when you multiply the avg with number of terms (n), you get the sum.

HTH
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25 Jan 2011, 02:49
dimitri92, thanks. I must have made a computational error.
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17 Feb 2011, 04:32
i know i read it somewhere, but i cant find it now. What are the ways to get the full Math book?
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30 May 2011, 10:25
 The general sum of a n term GP with common ratio r is given by

For sum of a GP:
When r > 1, the denominator is (r-1)
When r < 1, the denominator is (1-r)
Re: Math : Sequences & Progressions   [#permalink] 30 May 2011, 10:25

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