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

 It is currently 07 Dec 2013, 01:56

# Events & Promotions

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

# Math : Sequences & Progressions

Author Message
TAGS:
Manager
Joined: 08 Sep 2010
Posts: 174
Followers: 0

Kudos [?]: 11 [0], given: 18

Re: Math : Sequences & Progressions [#permalink]  05 Jun 2011, 12:06
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[highlight]& sufficient for a sequence to be an AP :[/highlight]
• \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}

[

Shouldn't the highlighted be GP and not AP?
_________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own.

If you like my explanations award kudos.

Manager
Joined: 08 Sep 2010
Posts: 174
Followers: 0

Kudos [?]: 11 [0], given: 18

Re: Math : Sequences & Progressions [#permalink]  09 Jun 2011, 20:55
I prefer the formula n/2{2a + (n-1)d}
_________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own.

If you like my explanations award kudos.

Intern
Joined: 14 Jun 2011
Posts: 5
Followers: 0

Kudos [?]: 1 [0], given: 0

Re: Math : Sequences & Progressions [#permalink]  21 Jun 2011, 01:44
can anyone pls specify the numbers of PS and DS practice ques for sequences and progressions available in official guide 12th edition....
Intern
Joined: 08 Jun 2011
Posts: 1
Followers: 0

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

Re: Math : Sequences & Progressions [#permalink]  23 Jun 2011, 14:23
One of the example given in GP is
"All positive odd and negative even numbers : {1,-2,3,-4,...}"

I don't think its a GP.

Regards,
Manager
Joined: 29 Jun 2011
Posts: 78
Followers: 3

Kudos [?]: 11 [0], given: 46

Re: Math : Sequences & Progressions [#permalink]  01 Jul 2011, 13:17
shrouded1 wrote:
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

I think i spotted a mistake...nothing of great consequence though...

"a" is not defined here...it should be "b"

correct me kindly if i am wrong

also thanks for the post...hugely informative and all-encompassing.
_________________

It matters not how strait the gate,
How charged with punishments the scroll,
I am the master of my fate :
I am the captain of my soul.
~ William Ernest Henley

Manager
Joined: 09 Feb 2012
Posts: 72
Location: India
Concentration: Marketing, Strategy
GMAT 1: 640 Q48 V31
GPA: 3.45
WE: Science (Health Care)
Followers: 1

Kudos [?]: 13 [0], given: 41

Re: Math : Sequences & Progressions [#permalink]  05 Mar 2012, 03:47
shrouded1 wrote:
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 [highlight]even.[/highlight] 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

Amazing material no doubt.

A small typo that I noticed, please check.

Thanks.
Manager
Joined: 09 Feb 2012
Posts: 72
Location: India
Concentration: Marketing, Strategy
GMAT 1: 640 Q48 V31
GPA: 3.45
WE: Science (Health Care)
Followers: 1

Kudos [?]: 13 [0], given: 41

Re: Math : Sequences & Progressions [#permalink]  05 Mar 2012, 03:54
Amazing material no doubt.

A small typo that I noticed, please check.

Thanks.[/quote]

Forgot to mention that the typo is in Misc Notes section.
I have highlighted it.

Thanks.
Manager
Joined: 09 Feb 2012
Posts: 72
Location: India
Concentration: Marketing, Strategy
GMAT 1: 640 Q48 V31
GPA: 3.45
WE: Science (Health Care)
Followers: 1

Kudos [?]: 13 [0], given: 41

Re: Math : Sequences & Progressions [#permalink]  05 Mar 2012, 04:14
Quote:
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

1023/1024 is very close to 1, so this sum is very close to 1/3

Can you please elaborate this, I am not able to deduce the outcome.
Math Expert
Joined: 02 Sep 2009
Posts: 15073
Followers: 2519

Kudos [?]: 15480 [0], given: 1552

Re: Math : Sequences & Progressions [#permalink]  05 Mar 2012, 06:03
Expert's post
pratikbais wrote:
Quote:
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

1023/1024 is very close to 1, so this sum is very close to 1/3

Can you please elaborate this, I am not able to deduce the outcome.

Discussed here: for-every-integer-k-from-1-to-10-inclusive-the-kth-term-of-88874.html
Similar question: for-every-integer-m-from-1-to-100-inclusive-the-mth-term-128575.html

Hope it helps.
_________________
Manager
Joined: 23 Feb 2012
Posts: 216
Location: India
Concentration: Finance, Entrepreneurship
Schools: Said
GMAT 1: 710 Q44 V44
GPA: 2.9
WE: Marketing (Computer Software)
Followers: 2

Kudos [?]: 35 [0], given: 22

Re: Math : Sequences & Progressions [#permalink]  07 Mar 2012, 04:17
How often do GP and HP come in the GMAT? I can believe AP is certainly tested. But are GP and HP in the same class as probability and combinatorics which appear only at the 750-800 range?
_________________

If you like it, Kudo it!

"There is no alternative to hard work. If you don't do it now, you'll probably have to do it later. If you didn't need it now, you probably did it earlier. But there is no escaping it."

710 Debrief. Crash and Burn

Math Expert
Joined: 02 Sep 2009
Posts: 15073
Followers: 2519

Kudos [?]: 15480 [0], given: 1552

Re: Math : Sequences & Progressions [#permalink]  07 Mar 2012, 05:31
Expert's post
budablasta wrote:
How often do GP and HP come in the GMAT? I can believe AP is certainly tested. But are GP and HP in the same class as probability and combinatorics which appear only at the 750-800 range?

There are some questions from GMAT Prep for which knowing the properties of GP might be useful (check the post above yours for an example). Though I've never seen the GMAT question testing/mentioning HP.
_________________
Manager
Joined: 23 Feb 2012
Posts: 216
Location: India
Concentration: Finance, Entrepreneurship
Schools: Said
GMAT 1: 710 Q44 V44
GPA: 2.9
WE: Marketing (Computer Software)
Followers: 2

Kudos [?]: 35 [0], given: 22

Re: Math : Sequences & Progressions [#permalink]  07 Mar 2012, 06:08
Thank you Bunuel. Just wanted to check before going into the relatively arcane stuff!
_________________

If you like it, Kudo it!

"There is no alternative to hard work. If you don't do it now, you'll probably have to do it later. If you didn't need it now, you probably did it earlier. But there is no escaping it."

710 Debrief. Crash and Burn

Senior Manager
Joined: 16 Feb 2012
Posts: 263
Concentration: Finance, Economics
Followers: 4

Kudos [?]: 31 [0], given: 102

Re: Math : Sequences & Progressions [#permalink]  03 May 2012, 04:11
[quote="shrouded1"]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 [highlight]b_1*\frac{r^n - 1}{r-1}[/highlight]
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
[highlight]S=b\frac{1-r^n}{1-r}[/highlight]=\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

[/quote

Something that should be the same formula is different at different places. Explain!
b_1*\frac{r^n - 1}{r-1}
and S=b\frac{1-r^n}{1-r}
_________________

Kudos if you like the post!

Failing to plan is planning to fail.

Manager
Status: Re-take.. The OG just loves me too much.
Joined: 18 Jun 2012
Posts: 67
Location: India
GMAT 1: 600 Q44 V29
WE: Information Technology (Consulting)
Followers: 2

Kudos [?]: 11 [0], given: 48

Re: Math : Sequences & Progressions [#permalink]  04 Sep 2012, 07:21
Hi,
Can some mod take a look into the example 2 provided under Geometric Progression?.
It states series - { 1, -2, 3, -4...} as an example with b1 = 1 and r = -1...

However, with the G.P formula, it looks incorrect as the common ratio thing doesn't hold true.

Kindly help.
_________________

Live Life the Way YOU Love It !!

GmatPrep1 [10/09/2012] : 650 (Q42;V38) - need to make lesser silly mistakes.
MGMAT 1 [11/09/2012] : 640 (Q44;V34) - need to improve quant pacing and overcome verbal fatigue.

Manager
Status: K... M. G...
Joined: 22 Oct 2012
Posts: 52
GMAT Date: 08-27-2013
GPA: 3.8
Followers: 0

Kudos [?]: 5 [0], given: 118

Re: Math : Sequences & Progressions [#permalink]  03 Nov 2012, 10:49
shrouded1 wrote:
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

its such a nice explanation....

I would like to understand more on the points which i have marked in RED.[a_4=a_1+3d, a_12=a_1+11d]. Please help me understanding
Math Expert
Joined: 02 Sep 2009
Posts: 15073
Followers: 2519

Kudos [?]: 15480 [0], given: 1552

Re: Math : Sequences & Progressions [#permalink]  10 Jul 2013, 23:07
Expert's post
Bumping for review*.

*New project from GMAT Club!!! Check HERE

_________________
Re: Math : Sequences & Progressions   [#permalink] 10 Jul 2013, 23:07
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