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Math : Sequences & Progressions

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Re: Math : Sequences & Progressions  [#permalink]

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New post 30 Jun 2014, 10:13
f an infinite GP is summable (|r|<1) then the sum is \frac{b_1}{1-r}

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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New post 30 Jun 2014, 10:53
sagnik2422 wrote:
f an infinite GP is summable (|r|<1) then the sum is \frac{b_1}{1-r}

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 \(sum=\frac{b}{1-r}\), 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 \(sum=\frac{b}{1-r}=\frac{\frac{1}{2}}{1-\frac{1}{3}}=\frac{3}{4}\).

Questions to practice:
for-every-integer-m-from-1-to-100-inclusive-the-mth-term-128575.html
for-every-integer-k-from-1-to-10-inclusive-the-kth-term-of-88874.html
a-square-is-drawn-by-joining-the-midpoints-of-the-sides-of-a-102880.html
m17q5-72268.html
ax-y-is-an-operation-that-adds-1-to-y-135277.html

Hope it helps.
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New post 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 \(b_1*\frac{r^n - 1}{r-1}\)
If an infinite GP is summable (|r|<1) then the sum is \(\frac{b_1}{1-r}\)
"

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
\(S=b_1\frac{1-r^n}{1-r}\)
not

\(b_1*\frac{r^n - 1}{r-1}\)
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New post 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 \(b_1*\frac{r^n - 1}{r-1}\)
If an infinite GP is summable (|r|<1) then the sum is \(\frac{b_1}{1-r}\)
"

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
\(S=b_1\frac{1-r^n}{1-r}\)
not

\(b_1*\frac{r^n - 1}{r-1}\)


There is no typo there. Those two are the same: factor -1 from denominator and numerator and reduce.
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New post 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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New post 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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New post 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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New post 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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New post 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. :-D
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Re: Math : Sequences & Progressions  [#permalink]

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New post 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 :think:

Kindly correct the typo. Bunuel Thank you very much for the post.
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New post 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 :think:

Kindly correct the typo. Bunuel Thank you very much for the post.


Edited. Thank you.
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New post 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*(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





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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New post 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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New post 10 Dec 2017, 06:09
1
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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New post 10 Dec 2017, 06:49
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New post 28 Feb 2018, 11:00
shrouded1 wrote:
Sequences & Progressions
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created by: shrouded1



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Definition

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
  1. All odd positive integers : {1,3,5,7,...} \(a_1=1, d=2\)
  2. All positive multiples of 23 : {23,46,69,92,...} \(a_1=23, d=23\)
  3. 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 ? :)


\(a_n = a_{n-1} + d = a_1 + (n-1)d\) so \(a_{n-1}\) is the first term and it is the same as \(a_1\) ? \(a_n = a_{n-1} + d = a_1 + (n-1)d\)

\(a_i\) 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 Properties
Each of the following is necessary & sufficient for a sequence to be an AP :
  • \(a_i - a_{i-1} =\) 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 : \(\frac{a_i - a_k}{i-k} = \frac{a_j-a_k}{j-k}\)
<--- what does it mean ? could some give an example with real numbers:)
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New post 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^(n-1)
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^(n-1) -- 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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New post 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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New post 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
\(S=b\frac{1-r^n}{1-r}=\frac{1}{2} * \frac{1-(-1/2)^{10}}{1-(-1/2)} = \frac{1}{3} * \frac{1023}{1024}\)
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)


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New post 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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Re: Math : Sequences & Progressions   [#permalink] 06 May 2020, 06:33

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