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Re: Math : Sequences & Progressions
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05 Jun 2011, 12:06
DefinitionIt is a special type of sequence in which the ratio of consequetive terms is constant General Term\(b_n = b_{n1} * r = a_1 * r^{n1}\) \(b_i\) is the ith term \(r\) is the common ratio \(b_1\) is the first term Defining PropertiesEach of the following is necessary[highlight]& sufficient for a sequence to be an AP :[/highlight]  \(\frac{b_i}{b_{i1}} =\) 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 : \((\frac{b_i}{b_k})^{jk} = (\frac{b_j}{b_k})^{ik}\)
[ Shouldn't the highlighted be GP and not AP?



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Re: Math : Sequences & Progressions
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09 Jun 2011, 20:55
I prefer the formula n/2{2a + (n1)d}



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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....



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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.
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01 Jul 2011, 13:17
shrouded1 wrote: General Term b_n = b_{n1} * r = a_1 * r^{n1} 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 allencompassing.



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05 Mar 2012, 03:47
shrouded1 wrote: 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\(a_n = a_{n1} + d = a_1 + (n1)d\) \(a_i\) 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 :  \(a_i  a_{i1} =\) Constant
 If you pick any 3 consecutive terms, the middle one is the mean of the other two
 For all i,j > k >= 1 : \(\frac{a_i  a_k}{ik} = \frac{a_ja_k}{jk}\)
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 \(\frac{n}{2}(2a+(n1)d)\) The sum formula may be rewritten 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\(b_n = b_{n1} * r = a_1 * r^{n1}\) \(b_i\) 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 :  \(\frac{b_i}{b_{i1}} =\) 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 : \((\frac{b_i}{b_k})^{jk} = (\frac{b_j}{b_k})^{ik}\)
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 \(b_1*\frac{r^n  1}{r1}\) If an infinite GP is summable (r<1) then the sum is \(\frac{b_1}{1r}\) 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,...} \(b_1=1/4, r=1/4, sum=\frac{1/4}{(11/4)}=(1/3)\)
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 nonnegative 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*(n1)=2n1 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^(n1) 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 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 = \(0.5 * \frac{10.5^5}{10.5} = \frac{1}{2} * \frac{\frac{31}{32}}{\frac{1}{2}} = \frac{31}{32}\) Example 2In an arithmetic progression a1,a2,...,a22,a23, the common difference is nonzero, 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 nonzero. 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 \(S=b\frac{1r^n}{1r}=\frac{1}{2} * \frac{1(1/2)^{10}}{1(1/2)} = \frac{1}{3} * \frac{1023}{1024}\) 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 : \(\frac{15}{2} (2a_1 + (151)d) = \frac{15}{2} * (2a_1+14d) = 150\) Answer is (c) Additional Exercises Amazing material no doubt. A small typo that I noticed, please check. Thanks.



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Re: Math : Sequences & Progressions
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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.



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Re: Math : Sequences & Progressions
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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 Answer is d Can you please elaborate this, I am not able to deduce the outcome. Thanks in advance!



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Re: Math : Sequences & Progressions
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05 Mar 2012, 06:03
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 Answer is d Can you please elaborate this, I am not able to deduce the outcome. Thanks in advance! Discussed here: foreveryintegerkfrom1to10inclusivethekthtermof88874.htmlSimilar question: foreveryintegermfrom1to100inclusivethemthterm128575.htmlHope it helps.
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Re: Math : Sequences & Progressions
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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 750800 range?



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07 Mar 2012, 05:31
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 750800 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.
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07 Mar 2012, 06:08
Thank you Bunuel. Just wanted to check before going into the relatively arcane stuff!



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Re: Math : Sequences & Progressions
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03 May 2012, 04:11
[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\(a_n = a_{n1} + d = a_1 + (n1)d\) \(a_i\) 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 :  \(a_i  a_{i1} =\) Constant
 If you pick any 3 consecutive terms, the middle one is the mean of the other two
 For all i,j > k >= 1 : \(\frac{a_i  a_k}{ik} = \frac{a_ja_k}{jk}\)
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 \(\frac{n}{2}(2a+(n1)d)\) The sum formula may be rewritten 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\(b_n = b_{n1} * r = a_1 * r^{n1}\) \(b_i\) 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 :  \(\frac{b_i}{b_{i1}} =\) 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 : \((\frac{b_i}{b_k})^{jk} = (\frac{b_j}{b_k})^{ik}\)
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]\(b_1*\frac{r^n  1}{r1}\)[/highlight] If an infinite GP is summable (r<1) then the sum is \(\frac{b_1}{1r}\) 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,...} \(b_1=1/4, r=1/4, sum=\frac{1/4}{(11/4)}=(1/3)\)
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 nonnegative 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*(n1)=2n1 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^(n1) 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 = \(0.5 * \frac{10.5^5}{10.5} = \frac{1}{2} * \frac{\frac{31}{32}}{\frac{1}{2}} = \frac{31}{32}\) Example 2In an arithmetic progression a1,a2,...,a22,a23, the common difference is nonzero, 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 nonzero. 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 \([highlight]S=b\frac{1r^n}{1r}[/highlight]=\frac{1}{2} * \frac{1(1/2)^{10}}{1(1/2)} = \frac{1}{3} * \frac{1023}{1024}\) 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 : \(\frac{15}{2} (2a_1 + (151)d) = \frac{15}{2} * (2a_1+14d) = 150\) Answer is (c) Additional Exercises [/quote Something that should be the same formula is different at different places. Explain! \(b_1*\frac{r^n  1}{r1}\) and \(S=b\frac{1r^n}{1r}\) Make up your mind!



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Re: Math : Sequences & Progressions
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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.



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Re: Math : Sequences & Progressions
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03 Nov 2012, 10:49
shrouded1 wrote: 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\(a_n = a_{n1} + d = a_1 + (n1)d\) \(a_i\) 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 :  \(a_i  a_{i1} =\) Constant
 If you pick any 3 consecutive terms, the middle one is the mean of the other two
 For all i,j > k >= 1 : \(\frac{a_i  a_k}{ik} = \frac{a_ja_k}{jk}\)
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 \(\frac{n}{2}(2a+(n1)d)\) The sum formula may be rewritten 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\(b_n = b_{n1} * r = a_1 * r^{n1}\) \(b_i\) 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 :  \(\frac{b_i}{b_{i1}} =\) 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 : \((\frac{b_i}{b_k})^{jk} = (\frac{b_j}{b_k})^{ik}\)
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 \(b_1*\frac{r^n  1}{r1}\) If an infinite GP is summable (r<1) then the sum is \(\frac{b_1}{1r}\) 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,...} \(b_1=1/4, r=1/4, sum=\frac{1/4}{(11/4)}=(1/3)\)
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 nonnegative 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*(n1)=2n1 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^(n1) 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 = \(0.5 * \frac{10.5^5}{10.5} = \frac{1}{2} * \frac{\frac{31}{32}}{\frac{1}{2}} = \frac{31}{32}\) Example 2In an arithmetic progression a1,a2,...,a22,a23, the common difference is nonzero, 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 nonzero. 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 \(S=b\frac{1r^n}{1r}=\frac{1}{2} * \frac{1(1/2)^{10}}{1(1/2)} = \frac{1}{3} * \frac{1023}{1024}\) 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 : \(\frac{15}{2} (2a_1 + (151)d) = \frac{15}{2} * (2a_1+14d) = 150\) Answer is (c) Additional Exercises 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



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10 Jul 2013, 23:07
Bumping for review*. *New project from GMAT Club!!! Check HERE
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04 Mar 2014, 06:24
Hi! Please help me. I don't understand example 4. How did you deduce that 1/2 is the 1st term and ratio??



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04 Mar 2014, 07:52
luisagonz8888 wrote: Hi! Please help me. I don't understand example 4. How did you deduce that 1/2 is the 1st term and ratio?? Hi, This question is discussed here: foreveryintegerkfrom1to10inclusivethekthtermof88874.htmlSimilar question: foreveryintegermfrom1to100inclusivethemthterm128575.htmlHope it helps.
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01 May 2014, 15:04
In the Geometric Progression example #2, I don't understand how the sequence for the 'b_1=1' and 'r=1' you provided results in a sequence of 1,2,3,4,5 I think it just goes 1,1,1,1,1,..... Could someone please clarify?
Thanks



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02 May 2014, 01:49
thingamaraj wrote: In the Geometric Progression example #2, I don't understand how the sequence for the 'b_1=1' and 'r=1' you provided results in a sequence of 1,2,3,4,5 I think it just goes 1,1,1,1,1,..... Could someone please clarify?
Thanks You are correct. Removed that from the topic.
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