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Re: Math : Sequences & Progressions
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30 Jun 2014, 11:13
f an infinite GP is summable (r<1) then the sum is \frac{b_1}{1r}
can someone please explain what these means with numbers?
i guess what i'm asking for here is a question where we would use this concept



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Re: Math : Sequences & Progressions
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30 Jun 2014, 11:53
sagnik2422 wrote: f an infinite GP is summable (r<1) then the sum is \frac{b_1}{1r}
can someone please explain what these means with numbers?
i guess what i'm asking for here is a question where we would use this concept The sum of infinite geometric progression with common ratio \(r<1\), is \(sum=\frac{b}{1r}\), where \(b\) is the first term.For example, the sum of infinite geometric progression 1/2, 1/6, 1/18, 1/54, ... (first term = 1/2, common ratio = 1/3) is \(sum=\frac{b}{1r}=\frac{\frac{1}{2}}{1\frac{1}{3}}=\frac{3}{4}\). Questions to practice: foreveryintegermfrom1to100inclusivethemthterm128575.htmlforeveryintegerkfrom1to10inclusivethekthtermof88874.htmlasquareisdrawnbyjoiningthemidpointsofthesidesofa102880.htmlm17q572268.htmlaxyisanoperationthatadds1toy135277.htmlHope it helps.
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Re: Math : Sequences & Progressions
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09 Aug 2014, 20:24
"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 \(b_1*\frac{r^n  1}{r1}\) If an infinite GP is summable (r<1) then the sum is \(\frac{b_1}{1r}\) "
Awesome post! After 4 years, I still want to read your post. Just need to correct the typo I hightlight than your post is perfect. The correct formula is \(S=b_1\frac{1r^n}{1r}\) not \(b_1*\frac{r^n  1}{r1}\)



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Re: Math : Sequences & Progressions
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12 Aug 2014, 01:43



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Re: Math : Sequences & Progressions
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26 Nov 2014, 03:56
One small correction as highlighted in bold: 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 ODD. In either case this is also equal to the mean of the first and last terms first and last terms



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Re: Math : Sequences & Progressions
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24 Jul 2015, 03:03
Hello,
In "summation" for the arithmetic progression, where you give the general sum of a n term AP with common difference d, you have "a" instead of "a1", right?



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Re: Math : Sequences & Progressions
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29 Dec 2016, 22:45
Correction please.
..."n 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 odd. In either case this is also equal to the mean of the first and last terms"



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Re: Math : Sequences & Progressions
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18 Apr 2017, 01:21
Are geometric and harmonic progressions tested/ important for GMAT?
I have just studied arithmetic progressions till now.



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Re: Math : Sequences & Progressions
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18 Apr 2017, 01:24
Shiv2016 wrote: Are geometric and harmonic progressions tested/ important for GMAT?
I have just studied arithmetic progressions till now. I would suggest have the basic knowledge of these as well. You may encounter a question on these if you are on your way to Q51.
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Re: Math : Sequences & Progressions
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19 Aug 2017, 12:00
In the Geometric Progressions section under the Defining Properties Each of the following is necessary & sufficient for a sequence to be an AP : .. It should be GP. A similar typo is repeated in the Harmonic Progressions section under the Defining Properties. Must be a ctrl c + ctrl v issue Kindly correct the typo. Bunuel Thank you very much for the post.



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Re: Math : Sequences & Progressions
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Re: Math : Sequences & Progressions
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30 Aug 2017, 23:49
Quote: 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*(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 GP
For 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 terms
I am sure you mean that " 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 ODD. In either case this is also equal to the mean of the first and last terms " Thank you so much for the post though, is very valuable



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Re: Math : Sequences & Progressions
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25 Nov 2017, 22:36
Very helpful post indeed. I have studied lots of concepts using different prep sources and I must say what I learn on here on the forum is priceless. Sometimes I feel like what the heck did I study or how did I miss that such an elegant solution exists. Many thanks!
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Re: Math : Sequences & Progressions
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10 Dec 2017, 07:09
Minor correction: It should be "a finite AP, the mean of all the terms is also equal to the mean of the middle two terms if n is even and the middle term if n is odd."



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Re: Math : Sequences & Progressions
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Re: Math : Sequences & Progressions
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28 Feb 2018, 12:00
shrouded1 wrote: Sequences & ProgressionsThis post is a part of [ GMAT MATH BOOK] created by: shrouded1 Get The Official GMAT Club's App  GMAT TOOLKIT 2. The only app you need to get 700+ score! [ iOS App] [ Android App]  DefinitionSequence : 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,...} \(a_1=0.1, d=1\)
Bunuel can you help me with some questions. it is great post just lacks some examples in order to grasp the basic concept / more detailed explanation... can you please answer my questions below in red ? \(a_n = a_{n1} + d = a_1 + (n1)d\) so \(a_{n1}\) is the first term and it is the same as \(a_1\) ? \(a_n = a_{n1} + d = a_1 + (n1)d\) \(a_i\) is the ith term (where do you guys see \(a_i\) / ith term in the above formula? \(d\) 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 < what does it mean ? could some give an example with real numbers :)
 If you pick any 3 consecutive terms, the middle one is the mean of the other two
 For all i,j > k >= 1 : \(\frac{a_i  a_k}{ik} = \frac{a_ja_k}{jk}\)
< what does it mean ? could some give an example with real numbers:)



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Re: Math : Sequences & Progressions
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20 Aug 2018, 20:55
Consider the GP with b1=1,r=2b1=1,r=2 {1,2,4,8,16,32,...}, so b_n=2^(n1) Pick out the subsequence of terms b2,b4,b6,... New sequence is {4,16,64,...} which is a GP with b1=4 and r=4
Can someone explain to me how did we get the subsequence as 4,16,64...
b_n=2^(n1)  so b_2 would be 2 right? b_4 = 8
So sequence would be 2,8,32...am i missing something?




Re: Math : Sequences & Progressions &nbs
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