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guide to series and sequences... arithmetic and geometric

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Re: guide to series and sequences... arithmetic and geometric [#permalink]

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New post 24 Mar 2011, 22:43
Superb post!!!+1 to u...
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Re: guide to series and sequences... arithmetic and geometric [#permalink]

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New post 08 Apr 2011, 15:54
Seeing this very late, yet u still deserves a Kudos. Great job !!!
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Re: guide to series and sequences... arithmetic and geometric [#permalink]

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New post 10 Apr 2011, 03:03
. Thanks, it is nice and precise
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Re: guide to series and sequences... arithmetic and geometric [#permalink]

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New post 15 Jan 2012, 06:57
perfect explanations benjiboo.kudos
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Re: guide to series and sequences... arithmetic and geometric [#permalink]

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New post 28 Jun 2013, 07:31
would like to add one more Formula to the list above from one of the replies from Bunuel.

For geometric progression with common ratio |r|<1, the sum of the progression is sum=\frac{b}{1-r}, where b is the first term.

reference: a-square-is-drawn-by-joining-the-midpoints-of-the-sides-of-a-102880.html
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Re: guide to series and sequences... arithmetic and geometric [#permalink]

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New post 28 Nov 2013, 05:23
Thank you a lot for this guide!
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Re: guide to series and sequences... arithmetic and geometric [#permalink]

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New post 31 Jul 2014, 08:19
Kudos many years after the original post. :)
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Re: guide to series and sequences... arithmetic and geometric [#permalink]

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New post 05 Dec 2014, 09:51
GMATnickname wrote:
Great post... To follow up on the extension into consecutive integers that Sayysong initiated above: I would invite you to think about the Data Sufficiency problem below. If you know how to solve this problem, then you know how to extract the maximum information possible out of the fact that you are dealing with consecutive numbers.

x is an integer; Is x^3-x divisible by 24?
(1) x is even
(2) x is odd

The answer is (B). I am happy to explain.
Enjoy!


The statement (2) is not sufficient when x=1.
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guide to series and sequences... arithmetic and geometric [#permalink]

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New post 09 Mar 2015, 15:48
Hi - can you please advise what is the formula for an arithmetic sequence in which each term after the first is equal to the sum of the preceding term? In other words, a sequence in which we are not adding fixed amount to get the next term.
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Re: guide to series and sequences... arithmetic and geometric [#permalink]

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New post 09 Mar 2015, 22:29
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Hi mawus,

You're actually describing 2 DIFFERENT sequences. By definition, an arithmetic sequence is one that involves a CONSTANT.....constantly adding (or subtracting) a number throughout a sequence. Here are some examples:

1, 2, 3, 4, 5.....
2, 4, 6, 8, 10....
-3, 0, 3, 6, 9....
0, 1/2, 1, 3/2, 2, 5/2, 3.....

Sequences that are based on prior terms (without a constant) can come in a variety of 'forms.' For example, the classic Fibonacci sequence adds the prior 2 terms to get the 'next' term:

1, 1, 2, 3, 5, 8, 13, 21

1+1 = 2
1+2 = 3
2+3 = 5
3+5 = 8
Etc.

GMAT sequence questions are relatively rare (you'll likely see just 1 that has "sequence notation"), but they DO typically involve a 'formula' of some kind. Do you have a specific example in mind that you'd like to discuss?

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Re: guide to series and sequences... arithmetic and geometric [#permalink]

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New post 15 Feb 2016, 01:42
Thanks for the great summary! Just a few follow up questions:


"The product of n consecutive integers is always divisible by n! So, 4x5x6 (4*5*6=120) is divisible by 3!"
- This is true because I have tried a few examples, but conceptually can someone please explain why this works??

"If you have an odd number of terms in consecutive set, the sum of those numbers is divisible by the number of terms. This does not hold true for consecutive sets with an even amount of terms."
- Again, can someone please explain why this is true conceptually? It's hard for me (and I think a lot of others) to memorize rules without understanding why they are true.

"If the arithmetic mean of 3 consecutive integers is odd, the product of them is divisible by 8"
- Similarly, can someone explain why this is true (other than showing by examples)?

Thanks!
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Re: guide to series and sequences... arithmetic and geometric [#permalink]

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New post 17 Feb 2016, 21:01
happyface101 wrote:
Thanks for the great summary! Just a few follow up questions:


"The product of n consecutive integers is always divisible by n! So, 4x5x6 (4*5*6=120) is divisible by 3!"
- This is true because I have tried a few examples, but conceptually can someone please explain why this works??

"If you have an odd number of terms in consecutive set, the sum of those numbers is divisible by the number of terms. This does not hold true for consecutive sets with an even amount of terms."
- Again, can someone please explain why this is true conceptually? It's hard for me (and I think a lot of others) to memorize rules without understanding why they are true.

"If the arithmetic mean of 3 consecutive integers is odd, the product of them is divisible by 8"
- Similarly, can someone explain why this is true (other than showing by examples)?

Thanks!



Anyone out there? Justin Bieber!
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Re: guide to series and sequences... arithmetic and geometric [#permalink]

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New post 05 Mar 2016, 00:25
awesome post! bump bump bump
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Re: guide to series and sequences... arithmetic and geometric [#permalink]

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New post 05 Mar 2016, 01:33
Expert's post
happyface101 wrote:
happyface101 wrote:
Thanks for the great summary! Just a few follow up questions:


"The product of n consecutive integers is always divisible by n! So, 4x5x6 (4*5*6=120) is divisible by 3!"
- This is true because I have tried a few examples, but conceptually can someone please explain why this works??

"If you have an odd number of terms in consecutive set, the sum of those numbers is divisible by the number of terms. This does not hold true for consecutive sets with an even amount of terms."
- Again, can someone please explain why this is true conceptually? It's hard for me (and I think a lot of others) to memorize rules without understanding why they are true.

"If the arithmetic mean of 3 consecutive integers is odd, the product of them is divisible by 8"
- Similarly, can someone explain why this is true (other than showing by examples)?

Thanks!



Anyone out there? Justin Bieber!


Hi,
Just saw three of your Queries..
lets take them one by one..

Quote:
"The product of n consecutive integers is always divisible by n! So, 4x5x6 (4*5*6=120) is divisible by 3!"
- This is true because I have tried a few examples, but conceptually can someone please explain why this works??

n would repeat itself as some multiple, as every nth term, n-1 term will repeat as some multiple, as every (n-1)th term,.. and so on till 2 will repeat as a multiple of 2, as every 2nd term..
so If there are n consecutive terms, there will be atleast one multiple of n,n-1,...,2 in those consecutive terms..

Quote:
"If you have an odd number of terms in consecutive set, the sum of those numbers is divisible by the number of terms. This does not hold true for consecutive sets with an even amount of terms."
- Again, can someone please explain why this is true conceptually? It's hard for me (and I think a lot of others) to memorize rules without understanding why they are true.


if the numbers are consecutive, the sum is the number of terms, n, multipplied by the average..
if odd numbers are there, the average/mean will be the central value and this value multiplied by n will give you the SUM..
so SUM= central value *number of terms.. so central value = SUM/n
and central value is an integer
SAY numbers are a,b,c,d,e..
c is the central value and the average..
so a+b+c+d+e= 5c...
5c will always de div by number of terms 5..

Now if the number of terms are even, there are two central values and the average is between these two, that is, SUM of these two central value divided by 2..
say a,b,c,d...
average =(b+c)/2 and number of terms = 4..
so SUM = (b+c)/2 *4= 2(b+c)..
here we can not be SURE that 2(b+c) will be div by number of terms, 4...
if b and c are both odd or both even, ans is YES,
OTherwise NO..

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New post 05 Mar 2016, 01:40
Expert's post
hi,
Quote:
"If the arithmetic mean of 3 consecutive integers is odd, the product of them is divisible by 8"
- Similarly, can someone explain why this is true (other than showing by examples)?
..
Sorry posted the previous post w/o your 3 query..

since the middle number is ODD, both of the numbers on its either side will be even and one out of two will be atleast div by 4..
so their product will have a multiple of 2, amultiple of 4 and a multiple of 3..
so it will be div by 8 and also 24..
let the middle number be 2a+1.. so numbers are 2a, 2a+1, 2a+2..
product= 2a*(2a+1)(2a+2)= 2a*(2a+1)*2*(a+1)= 4*a*(2a+1)*(a+1)=..
sice a and a+1 are consecutive numbers, any one of them will be even and one odd.. so 4*a*(2a+1)*(a+1) will be div by 4*2(these 2 comes from a or a+1)

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Re: guide to series and sequences... arithmetic and geometric [#permalink]

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New post 15 Mar 2016, 07:04
benjiboo wrote:
I want to make this more clear for people who stumble on this post in the future. The following is meant to help one understand the entire topic that this falls under.

-------------------
PART I
-------------------

Topic: Sequences and Series

There are two types of sequences and two types of series. They are geometric sequences and arithmetic sequences, and geometric series and arithmetic series.

Geometric sequence vs arithmetic sequence

An arithmetic sequence is a sequence of numbers where each new term after the first is formed by adding a fixed amount called the common difference to the previous term in the sequence.

Set A={1,2,3,4,5,6,7,8,9,10}
Set B={2,4,6,8,10,12,14}
Set C={3,8,13,18,23,28}

In 'set A', the common difference is the fixed amount of one. In 'set B' the common difference is the fixed amount of two, and in 'set C' the common difference is the fixed amount of five. As you most likely noticed already, the common difference is found by finding the difference between two consecutive terms within the sequence. For example, in 'set C', to find the common difference compute (8-3=5).

A geometric sequence on the other hand, is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed number called the common ratio.

Set D={2,4,8,16,32}
Set E={3,9,27,81}
Set F={5,10,20,40,80}

You might notice that the difference between consecutive numbers in the above three sets are not a fixed amount. For instance, in 'set F', the first two terms (5 and 10) have a smaller difference than the last two terms (40 and 80). Therefore, the above sets are geometric sequences. The difference between two consecutive numbers is therefor the common ratio. To find the common ratio you simply take the ratio one consecutive number to the one before it. In 'set F' this would be (10/5=2). Therefore, n 'set F' the common ratio is two. In 'set E' the common ratio is (27/9=3). In 'set' D' the common ratio is two (32/16=2).

Summary
The difference between the two types of sequences is that in arithmetic sequences the consecutive numbers in a set differ by a fixed amount known as the common difference whereas in a geometric sequence the consecutive numbers in a set differ by a fixed number known as the common ratio.

Sequence vs Series

This is quite simple. A sequence is a list of numbers. A series is created by adding terms in the sequence. There you go, now you know the difference. So if you take 'set A' and add the terms then you have an arithmetic series. If you take 'set D' and add the terms, then you have a geometric series.

Sequence: {1, 3, 5, 7, 9, …}
Series: {1+3+5+7+9+…}

What the GMAT could ask us to do with sequences and series and how to do it!

There is no limit to what the GMAT can ask you to find when dealing with series and sequences. Here are some examples of things you may be asked to find/do with them.

(1) The sum of numbers in a series (which can be asked in many tricky ways such as the sum of all the numbers, sum of just the even numbers, sum of just the odd numbers, sum of only the numbers which are multiples of 7, sum of the first 10 numbers, and many more tricky ways!)

(2) The nth term in a sequence

(3) How many integers are there in a sequence

Anyway, now that you get the point... lets give you the formulas that will allow you to answer any question regarding series and sequences. I will then show you how to use the formulas to answer some questions that might not be intuitive of non math geniuses.

Formula for geometric sequence (when there is a common ratio)
dark green means subscript

Recursive (to find just the next term):

an = an-1 * r

Explicit (to find any nth term):

an = a1 * \(r^n-^1\)

an = nth term
a1 = the first term
r = common ratio

In reality, you only need to know the explicit formula, because you can find any term with it. I only put the recursive formula for understanding.

Formula for arithmetic sequence (when there is a common difference)
dark green means subscript

recursive:

an = an-1 + d

Explicit:

an = a1 + (n-1)d

an = nth term
a1 = first term
d = common difference

Again, you only need to know the explicit formula, because you can find any term with it. I only put the recursive formula for understanding.

Formula for geometric series (when there is a common ratio)
dark green means subscript

Sn = a1\(\frac{(1-r^n)}{(1-r)}\)

Sn = Sum of first nth terms
a1 = first term
r = common ratio
n = nth term

Formula for arithmetic series (when there is a common ratio)
dark green means subscript

Sn = \(\frac{n}{2}\)(a1 + an)

or

Sn = \(\frac{n}{2}\)(First term + Last term)

The above two equations are the same (I put them in both ways because some prep programs teach "first + last" but it is important to see that in the first of the two, the last term is identified as an. Well what if you do not know the last term? Then you have to calculate it using the equation for the nth term (solving for an) of an arithmetic sequence which is listed above... or you can substitute the formula for an into the first one of these two by replacing an with what is equals and simplifying. You get the following:

Sn = \(\frac{n}{2}\)[2a + (n-1)d]

Sn = sum of the series
a1 = the first term
an = the nth term
n = the number of terms
d = the common difference

----------------------------
PART 2
----------------------------
Now that we know all this information, there are some important things that are understood as well to ensure that the formulas are used correctly.

How to find the number of integers in a set

(Last term - First term) + 1

*A mistake is that people will forget to add the 1. The number of terms between 3 and 10 is not 7, it is 8. A common mistake is that people will calculate (10-3=7)... but this is wrong. Remember, as Manhattan GMAT says, "Add one before you are done".

*Notice how I used the word "term" and not number. This is important because sometimes you don't always just put the first and last number you are given. For example, If you are asked to find the number of even integers between 1 and 30, you don't use the "first number" in the set. The first number is "1", which is odd, and we are only speaking about even numbers. Therefore, the first term is "2", not "1", even though the set or question might have stated "from 1-30". Same goes with the last term. There is another step needed to answer this question though.

Find number of odd integers (or even) in a set

(\(\frac{(Last term - First term)}{2}\) + 1

*If the question is to find the number of odd integers between 2 and 30, then your first term is 3, and your last term is 29. They must be odd to fit in the set you are asked to analyze.
*If the question is find the number of even integers between 3 and 29, then your first term is 4, and your last term is 28.

Find number of integers that are a multiple of a certain number in a set

GMAT questions can get tricky, but luckily not too tricky. For example... What if you are asked to "find the number of multiples of 7 between 2 and 120"?

(\(\frac{(Last term - First term)}{increment}\)) + 1

\([\frac{(119 - 7)}{7}]\)+ 1

All you have to do is instead of dividing our old formula by 2, you divide it by the increment. Also, notice how my first and last terms are the first term that is a multiple of 7 and the last term that is a multiple of seven within the set!

Sum of odd numbers in a series

This seems to be a popular topic on GMAT forums. Its quite simple. You already know everything you need to after reading this post. It is a two step problem. Here are the two steps:

(1) Find the number of odd terms. This is you "n" value now.
(2) Plug in the "n" value into the formula for an arithmetic series.

-------------------------
PART III
-------------------------

There are some short cuts and concepts that you should know about this topic.

(1) The mean and the medium of any arithmetic sequence is equal to the average of the first and last terms.
(2) The sum of an arthritic sequence is equal to the mean (average) times the number of terms.
(3) The product of n consecutive integers is always divisible by n! So, 4x5x6 (4*5*6=120) is divisible by 3!
(4) If you have an odd number of terms in consecutive set, the sum of those numbers is divisible by the number of terms.
(5) number four (above) does not hold true for consecutive sets with an even amount of terms.


Additional Exercises





------------------------
Thank you
Benjiboo




Wow. Great post!! Thanks so much.

Suggested edit:

Explicit (to find any nth term):

an = a1 * r^(n−1) ---- I see as (n_1)
Re: guide to series and sequences... arithmetic and geometric   [#permalink] 15 Mar 2016, 07:04

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