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Intern  Joined: 20 May 2014
Posts: 32
Re: Math : Sequences & Progressions  [#permalink]

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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
Math Expert V
Joined: 02 Sep 2009
Posts: 57155
Re: Math : Sequences & Progressions  [#permalink]

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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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Intern  Joined: 29 May 2014
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Re: Math : Sequences & Progressions  [#permalink]

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"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

$$b_1*\frac{r^n - 1}{r-1}$$
Math Expert V
Joined: 02 Sep 2009
Posts: 57155
Re: Math : Sequences & Progressions  [#permalink]

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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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Intern  Joined: 15 Mar 2013
Posts: 7
Re: Math : Sequences & Progressions  [#permalink]

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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
Senior Manager  Status: Math is psycho-logical
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Re: Math : Sequences & Progressions  [#permalink]

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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?
Intern  B
Joined: 13 Nov 2016
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Schools: ESSEC '20
GMAT 1: 700 Q48 V40 GPA: 3.9
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Re: Math : Sequences & Progressions  [#permalink]

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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"
Director  G
Joined: 02 Sep 2016
Posts: 655
Re: Math : Sequences & Progressions  [#permalink]

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Are geometric and harmonic progressions tested/ important for GMAT?

I have just studied arithmetic progressions till now.
Board of Directors V
Status: Stepping into my 10 years long dream
Joined: 18 Jul 2015
Posts: 3603
Re: Math : Sequences & Progressions  [#permalink]

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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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Manager  B
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Re: Math : Sequences & Progressions  [#permalink]

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 57155
Re: Math : Sequences & Progressions  [#permalink]

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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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Intern  B
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Re: Math : Sequences & Progressions  [#permalink]

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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 Manager  B
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GMAT 1: 640 Q43 V35 Re: Math : Sequences & Progressions  [#permalink]

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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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Intern  B
Joined: 03 Dec 2017
Posts: 1
Re: Math : Sequences & Progressions  [#permalink]

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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."
Math Expert V
Joined: 02 Sep 2009
Posts: 57155
Re: Math : Sequences & Progressions  [#permalink]

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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  [#permalink]

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shrouded1 wrote:
Sequences & Progressions This post is a part of [GMAT MATH BOOK]

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:)
Intern  B
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Re: Math : Sequences & Progressions  [#permalink]

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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?
Senior Manager  S
Joined: 12 Sep 2017
Posts: 298
Re: Math : Sequences & Progressions  [#permalink]

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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! Re: Math : Sequences & Progressions   [#permalink] 20 Jan 2019, 12:40

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