[quote="shrouded1"]
Sequences & ProgressionsThis post is a part of [
GMAT MATH BOOK]
created by: shrouded1DefinitionSequence : 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 ProgressionsDefinitionIt 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 PropertiesEach 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 : \(\frac{a_i - a_k}{i-k} = \frac{a_j-a_k}{j-k}\)
SummationThe 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 ProgressionsDefinitionIt 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 PropertiesEach 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 : \((\frac{b_i}{b_k})^{j-k} = (\frac{b_j}{b_k})^{i-k}\)
SummationThe 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- All positive powers of 2 : {1,2,4,8,...} \(b_1=1, r=2\)
- All positive odd and negative even numbers : {1,-2,3,-4,...} \(b_1=1, r=-1\)
- 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 ProgressionsDefinitionIt is a special type of sequence in which if you take the inverse of every term, this new sequence forms an AP
Important PropertiesOf 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 : LinkageEach 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 NotesA subsequence (any set of consequutive terms) of an AP is an APA subsequence (any set of consequutive terms) of a GP is a GPA subsequence (any set of consequutive terms) of a HP is a HPIf 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 APFor 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 GPFor 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 termsSome examplesExample 1A 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 ?
SolutionP(<=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 2In 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 3For positive integers a,b (a<b) arrange in ascending order the quantities a, b, sqrt(ab), avg(a,b), 2ab/(a+b)
SolutionUsing the inequality AM>=GM>=HM, the solution is :
a <= 2ab/(a+b) <= Sqrt(ab) <= Avg(a,b) <= b
Example 4For 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.
SolutionThe 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
Answer is d
Example 5The 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
[/quote
Something that should be the same formula is different at different places. Explain!
\(b_1*\frac{r^n - 1}{r-1}\)
and \(S=b\frac{1-r^n}{1-r}\)
Make up your mind!
close
05 Jun 2011, 13:06
