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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

All odd positive integers : {1,3,5,7,...} a_1=1, d=2

All positive multiples of 23 : {23,46,69,92,...} a_1=23, d=23

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

All positive powers of 2 : {1,2,4,8,...} b_1=1, r=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 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)

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
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Re: Math : Sequences & Progressions [#permalink]
29 Oct 2010, 20:19

Valuable resource. Kudos +1
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"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??

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.

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
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