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Can someone tell me the formula of consecutive sum?

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Can someone tell me the formula of consecutive sum? [#permalink] New post 14 Oct 2009, 13:51
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Like there are 200 numbers consecutive postive numbers and i need to find sum for it?

I also need formula for consecutive postive even, odd sums.

Thanks in advance.
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Re: Can someone tell me the formula of consecutive sum? [#permalink] New post 14 Oct 2009, 17:04
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Well, this is nothing but "Arithmetic Progression". To make it simple, I will tell you 3 formulas which you are looking for:

1) Sum of consecutive integers:
Lets say we want to find out the sum of first n positive integers. The formula for that is:

Sum = n*(n+1)/2

Examples:
sum of first 200 integers = 200*(201)/2 = 20100
sum of first 10 integers = 10*(11)/2 = 55

2) Sum of consecutive EVEN integers:
Lets say we want to find out the sum of consecutive positive even integers from 2 to n, where n is EVEN. The formula for that is:

Sum = n*(n+2)/4

Examples:
sum of even integers from 1 to 200 = 200*(202)/4 = 10100
sum of even integers from 1 to 10 = 10*(12)/4 = 30

3) Sum of consecutive ODD integers:
Lets say we want to find out the sum of consecutive positive odd integers from 1 to n, where n is ODD. The formula for that is:

Sum = (n+1)*(n+1)/4 = (n+1)^2/4

Examples:
sum of odd integers from 1 to 199 = 200*(200)/4 = 10000
sum of odd integers from 1 to 9 = 10*(10)/4 = 25

Hope this helps.
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Re: Can someone tell me the formula of consecutive sum? [#permalink] New post 14 Oct 2009, 18:43
Thank you very much!
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Re: Can someone tell me the formula of consecutive sum? [#permalink] New post 14 Oct 2009, 21:07
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An easier way to add n consecutive numbers is

n*(firstnumber + lastnumber)/2

so for the first 200 numbers it will be
200*(1+200)/2
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Re: Can someone tell me the formula of consecutive sum? [#permalink] New post 15 Oct 2009, 03:43
rohitbhotica wrote:
An easier way to add n consecutive numbers is

n*(firstnumber + lastnumber)/2

so for the first 200 numbers it will be
200*(1+200)/2


@rohitbhotica

The only thing to remember is, if first number is not 1, then you may not multiply by n. e.g. if you want to find the sum of consecutive numbers from 10 to 20, then 20*(10+20)/2 does not give correct results.
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Re: Can someone tell me the formula of consecutive sum? [#permalink] New post 15 Oct 2009, 03:52
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hgp2k wrote:
rohitbhotica wrote:
An easier way to add n consecutive numbers is

n*(firstnumber + lastnumber)/2

so for the first 200 numbers it will be
200*(1+200)/2


@rohitbhotica

The only thing to remember is, if first number is not 1, then you may not multiply by n. e.g. if you want to find the sum of consecutive numbers from 10 to 20, then 20*(10+20)/2 does not give correct results.



no in my explanation n is the number of numbers that are added. So a sum from 10 to 20 will have n as 11
so the answer will be 11*(10+20)/2 = 165 which is the answer.
I hope that helps
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Re: Can someone tell me the formula of consecutive sum? [#permalink] New post 15 Oct 2009, 05:46
got it :)
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Re: Can someone tell me the formula of consecutive sum? [#permalink] New post 15 Oct 2009, 07:05
and the best part about this is that it works for any number of numbers in a arithmetic progression..
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Re: Can someone tell me the formula of consecutive sum? [#permalink] New post 28 Oct 2009, 18:35
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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.

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

Last edited by benjiboo on 06 Nov 2009, 17:41, edited 1 time in total.
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Re: Can someone tell me the formula of consecutive sum? [#permalink] New post 29 Oct 2009, 12:14
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Re: Can someone tell me the formula of consecutive sum? [#permalink] New post 06 Nov 2009, 06:52
Well, that's great, Benjiboo. I understand almost all about geometric & arithmetic sequences now.

However, I still have one question about the Increment 4 above.

I suppose in the example about the number of multiples of 7 between 2 and 120, the increment should be 7. If so, we have

(119-7)/7 + 1 = 17 numbers --> this is true.

If increment is 4 --> (119-7)/4 + 1= 29 numbers --> not true.

Correct me if I am wrong.
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Re: Can someone tell me the formula of consecutive sum? [#permalink] New post 06 Nov 2009, 17:47
cheryl3007 wrote:
Well, that's great, Benjiboo. I understand almost all about geometric & arithmetic sequences now.

However, I still have one question about the Increment 4 above.

I suppose in the example about the number of multiples of 7 between 2 and 120, the increment should be 7. If so, we have

(119-7)/7 + 1 = 17 numbers --> this is true.

If increment is 4 --> (119-7)/4 + 1= 29 numbers --> not true.

Correct me if I am wrong.


Hi, sorry I had a typo in there, in which I had a 4 instead of a 7 in the equation. I have fixed it now.

As to clear up anyone who might now be confused by this post and my error, when you do such an equation for an increment, make sure that the FIRST TERM and LAST TERM are multiples of the increment. So if you wanted to find the multiples of 7 between 2 and 120, you would use 119 as the LAST TERM and 7 as the FIRST TERM. However, if you wanted to find the multiples of 4 between 2 and 120, your FIRST TERM would be 4, and your LAST TERM would be 120.

As for anyone else reading this, please PM me if you are having problems with this topic. I am in the middle of creating a GMAT MATH STRATEGY GUIDE that covers every GMAT topic for a quick reference. There are many different ways to think about each topic in math. For example, there is another way to think about series and sequences that I did not like here that may be easier for some people. My guide will include all of that, but it is not finished yet. However, please do PM me if you need help with any topic and nobody has answered your post or so on!

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Re: Can someone tell me the formula of consecutive sum? [#permalink] New post 25 Dec 2013, 06:29
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Re: Can someone tell me the formula of consecutive sum? [#permalink] New post 06 Jan 2014, 09:06
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Hi

Sum of n consecutive/first 'n' natural numbers = n(n+1)/2

Sum of n consecutive even numbers = n(n+1)

Sum of the squares of the first n natural numbers = n(n+1)(2n+1)/6

Sum of the cubes of first n natural numbers = [n(n+1)/2]^2 (the whole square)
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Re: Can someone tell me the formula of consecutive sum?   [#permalink] 06 Jan 2014, 09:06
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