Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 14 Mar 2014, 23:00

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Math : Sequences & Progressions

Author Message
TAGS:
Retired Moderator
Joined: 02 Sep 2010
Posts: 807
Location: London
Followers: 74

Kudos [?]: 385 [30] , given: 25

Math : Sequences & Progressions [#permalink]  28 Sep 2010, 15:32
30
KUDOS
Sequences & Progressions

This post is a part of [GMAT MATH BOOK]

created by: shrouded1

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 AP :
• \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 positive odd and negative even numbers : {1,-2,3,-4,...} b_1=1, r=-1
3. 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 AP

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 even. 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

_________________

Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

My GMAT story

Last edited by shrouded1 on 02 Nov 2010, 11:29, edited 8 times in total.
 Kaplan Promo Code Knewton GMAT Discount Codes Veritas Prep GMAT Discount Codes
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2792
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 155

Kudos [?]: 819 [1] , given: 232

Re: Math : Sequences & Progressions [#permalink]  28 Sep 2010, 16:25
1
KUDOS
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.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html

Intern
Joined: 26 Sep 2010
Posts: 16
Location: Singapore
WE 1: 6.5 years in Decision Sciences
Followers: 1

Kudos [?]: 0 [0], given: 0

Re: Math : Sequences & Progressions [#permalink]  28 Sep 2010, 21:33
Thanks for putting it here!
Retired Moderator
Joined: 02 Sep 2010
Posts: 807
Location: London
Followers: 74

Kudos [?]: 385 [2] , given: 25

Re: Math : Sequences & Progressions [#permalink]  29 Sep 2010, 09:27
2
KUDOS
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 ?
_________________

Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

My GMAT story

Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 413
Location: Milky way
Schools: ISB, Tepper - CMU, Chicago Booth, LSB
Followers: 6

Kudos [?]: 95 [0], given: 50

Re: Math : Sequences & Progressions [#permalink]  29 Sep 2010, 18:30
Great initiative. +1 to you.
_________________

Support GMAT Club by putting a GMAT Club badge on your blog

Manager
Joined: 16 Aug 2009
Posts: 222
Followers: 3

Kudos [?]: 14 [0], given: 18

Re: Math : Sequences & Progressions [#permalink]  30 Sep 2010, 04:25
Great one
Kudos to you !
Retired Moderator
Joined: 02 Sep 2010
Posts: 807
Location: London
Followers: 74

Kudos [?]: 385 [0], given: 25

Re: Math : Sequences & Progressions [#permalink]  07 Oct 2010, 00:33
added a couple more solved GMAT style questions to the end
_________________

Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

My GMAT story

Senior Manager
Joined: 31 Mar 2010
Posts: 415
Location: Europe
Followers: 2

Kudos [?]: 33 [0], given: 26

Re: Math : Sequences & Progressions [#permalink]  07 Oct 2010, 00:56
Good to know. It can definitely same valuable time.
Manager
Joined: 19 Apr 2010
Posts: 216
Schools: ISB, HEC, Said
Followers: 4

Kudos [?]: 14 [0], given: 28

Re: Math : Sequences & Progressions [#permalink]  17 Oct 2010, 23:10
This is great post but I was just wondering whether we need to know these concepts for GMAT
Retired Moderator
Joined: 02 Sep 2010
Posts: 807
Location: London
Followers: 74

Kudos [?]: 385 [0], given: 25

Re: Math : Sequences & Progressions [#permalink]  18 Oct 2010, 22:49
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
_________________

Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

My GMAT story

Senior Manager
Joined: 20 Jan 2010
Posts: 278
Schools: HBS, Stanford, Haas, Ross, Cornell, LBS, INSEAD, Oxford, IESE/IE
Followers: 11

Kudos [?]: 121 [0], given: 117

Re: Math : Sequences & Progressions [#permalink]  29 Oct 2010, 20:19
Valuable resource.
Kudos +1
_________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so."
Target=780
http://challengemba.blogspot.com
Kudos??

Intern
Joined: 14 Oct 2010
Posts: 1
Followers: 0

Kudos [?]: 3 [3] , given: 1

Re: Math : Sequences & Progressions [#permalink]  02 Nov 2010, 10:53
3
KUDOS
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.
Retired Moderator
Joined: 02 Sep 2010
Posts: 807
Location: London
Followers: 74

Kudos [?]: 385 [0], given: 25

Re: Math : Sequences & Progressions [#permalink]  02 Nov 2010, 11:28
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
_________________

Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

My GMAT story

Senior Manager
Joined: 08 Jun 2010
Posts: 454
Followers: 0

Kudos [?]: 25 [0], given: 39

Re: Math : Sequences & Progressions [#permalink]  11 Nov 2010, 20:52
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.
Retired Moderator
Joined: 02 Sep 2010
Posts: 807
Location: London
Followers: 74

Kudos [?]: 385 [0], given: 25

Re: Math : Sequences & Progressions [#permalink]  12 Nov 2010, 01:34
 ! 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
_________________

Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

My GMAT story

Director
Joined: 23 Apr 2010
Posts: 585
Followers: 2

Kudos [?]: 16 [0], given: 7

Re: Math : Sequences & Progressions [#permalink]  17 Jan 2011, 02:09

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

(lastterm - firstterm)*(lastterm - firstterm + 1)/2
Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 326
Followers: 10

Kudos [?]: 215 [0], given: 20

Re: Math : Sequences & Progressions [#permalink]  24 Jan 2011, 09:48
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

_________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Director
Joined: 23 Apr 2010
Posts: 585
Followers: 2

Kudos [?]: 16 [0], given: 7

Re: Math : Sequences & Progressions [#permalink]  25 Jan 2011, 01:49
dimitri92, thanks. I must have made a computational error.
Senior Manager
Joined: 08 Nov 2010
Posts: 424
Followers: 6

Kudos [?]: 28 [0], given: 161

Re: Math : Sequences & Progressions [#permalink]  17 Feb 2011, 03:32
i know i read it somewhere, but i cant find it now. What are the ways to get the full Math book?
Intern
Joined: 29 May 2011
Posts: 5
Schools: HBS
Followers: 0

Kudos [?]: 6 [0], given: 0

Re: Math : Sequences & Progressions [#permalink]  30 May 2011, 09:25
 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, 09:25
Similar topics Replies Last post
Similar
Topics:
Sequences 7 16 Jul 2005, 06:45
Progress 5 29 May 2006, 06:43
Progress? 6 05 Nov 2006, 18:13
Progression 1 25 Apr 2010, 11:22
Progress! 2 22 Mar 2012, 13:52
Display posts from previous: Sort by