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}

APs, GPs, HPs : Linkage

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
Answer is B

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
Answer is d

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
Answer is (c)

Additional Exercises

• toughest-progression-questions-99380.html
• arithmetic-progression-82035.html
• PS Sequence Questions
• DS Sequence questions

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Let me know what else ?
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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.

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

Great initiative. +1 to you.
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added a couple more solved GMAT style questions to the end
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This is great post but I was just wondering whether we need to know these concepts for GMAT

Valuable resource.
Kudos +1
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Kudos??

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.

Can someone please clarify:

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

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