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Re: Math : Sequences & Progressions
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30 Jun 2014, 10:13
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, 10:53
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, 19:24
"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, 00:43
linhntle wrote: "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 \(b_1*\frac{r^n  1}{r1}\) There is no typo there. Those two are the same: factor 1 from denominator and numerator and reduce.
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Re: Math : Sequences & Progressions
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26 Nov 2014, 02:56
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, 02:03
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, 21:45
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, 00:21
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, 00:24
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, 11:00
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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20 Aug 2017, 00:28
shoumkrish wrote: 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. Edited. Thank you.
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Re: Math : Sequences & Progressions
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30 Aug 2017, 22:49
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, 21:36
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, 06:09
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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10 Dec 2017, 06:49
Eudaimonia wrote: 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." ______________ Edited. Thank you.
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Re: Math : Sequences & Progressions
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28 Feb 2018, 11:00
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, 19:55
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?



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Re: Math : Sequences & Progressions
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20 Jan 2019, 11:40
Hello everyone!
Could someone please explain to me what is exactly the following value?
General Term bn=bn−1∗r=a1∗rn−1bn=bn−1∗r=a1∗rn−1 bibi is the ith term Where is that term in the formula? bn=bn−1∗r=a1∗rn−1bn=bn−1∗r=a1∗rn−1 rr is the common ratio b1b1 is the first term
Kind regards!



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Math : Sequences & Progressions
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05 May 2020, 19:37
Can someone please explain how to solve Example #4 and Example #5 in more detail? I am finding it confusing.shrouded1 wrote: 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 \(S=b\frac{1r^n}{1r}=\frac{1}{2} * \frac{1(1/2)^{10}}{1(1/2)} = \frac{1}{3} * \frac{1023}{1024}\) 1023/1024 is very close to 1, so this sum is very close to 1/3 Answer is d
Example 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 : \(\frac{15}{2} (2a_1 + (151)d) = \frac{15}{2} * (2a_1+14d) = 150\) Answer is (c)




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Re: Math : Sequences & Progressions
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06 May 2020, 06:33
Scratch that  I figured it out! leoxcvi wrote: Can someone please explain how to solve Example #4 and Example #5 in more detail? I am finding it confusing.shrouded1 wrote: 




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