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For any positive integer n, the sum of the first n positive integers

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For any positive integer n, the sum of the first n positive integers  [#permalink]

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For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

Originally posted by CharmWithSubstance on 11 May 2010, 15:57.
Last edited by Bunuel on 15 Mar 2019, 03:15, edited 1 time in total.
Updated.
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 19 Feb 2012, 23:11
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For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

Approach #1:

Even integer between 99 and 301 represent evenly spaced set (aka arithmetic progression): 100, 102, 104, ..., 300. Now, the sum of the elements in any evenly spaced set is the mean (average) multiplied by the number of terms. (Check Number Theory chapter of Math Book for more: http://gmatclub.com/forum/math-number-theory-88376.html)

Average of the set: (largest+smallest)/2=(300+100)/2=200;
# of terms: (largest-smallest)/2+1=(300-100)/2+1=101 (check this: http://gmatclub.com/forum/totally-basic ... ml#p730075);

The sum = 200*101= 20,200.

Answer: B.


Approach #2:

Using the formula of the sum of the first n positive integers: n(n+1)/2.

100+102+...+300=2(50+51+..+150). Now, the sum of the integers from 50 to 150, inclusive equals to the sum of the integers from 1 to 150, inclusive minus the sum of the integers from 1 to 49, inclusive. 2(50+51+..+150)=2*(150(150+1)/2-49(49+1)/2)=20,200.

Answer: B.

Hope it helps.
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 19 Feb 2012, 19:36
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For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

a = first term = 100
l = last term = 300

Total number of terms = ((300-100)/2 ) + 1 = 101

sum = n (a + l ) / 2 = 101 (100 + 300 ) / 2 = 20200
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For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 11 May 2010, 16:54
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CharmWithSubstance wrote:
For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150


In any set of consecutive integers the sum = average*number of integers

average = 200
total = 150-50 + 1 = 101

150 because 300 is the 150th even integer
50 because 100 is the 50th even integer

101*200 = 20200
Therefore there are 101 integers
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For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 11 May 2010, 20:26
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CharmWithSubstance wrote:
For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150


sum of all even integer 1 to 301 = 2 * (1+2+...+150) = (2*150*151)/2 = 150*151
sum of all even integer 1 to 99 = 2 * (1+2+....+49) = (2*49*50)/2 = 49*50

required sum = 150*151 - 49*50 = 50*(453 - 49) =
404 * 50 = 20200
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 10 Jul 2010, 18:14
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They are basically asking for the sum of even integers between/inclusive of 100 and 300.

Sum = \(100+102+ .... 300 = 2(50+51+52+ ....150)\) - We've converted even integers into consecutive integers for using this formula.

Now, let's just add and subtract 1 to 50 to this series, so we get:

\(2[(1+2+3 ..... 50+51+ ..... 150)-(1+2+3 ...50)]\)

So, we've broken this sum down into two sums of 1 .... x

Sum of 1 .... 150 = \(\frac{150(150+1)}{2} = 75*151\)

Similarly, sum of 1 .... 49 = \(\frac{49(49+1)}{2} = 25*49\)

So we get the answer to be: \(2[(75*151) - (25*49) = 20200\)

The reason why we do this is because in the sum of even integers, when 2 is factored out, you get consecutive integers as I've shown here. And since we have a formula for the first "n" integers and not for some random range of integers we try to reduce what we're asked to find to this form

Hope this answers your question.
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 11 Jul 2010, 06:36
thanks!! it makes sense now...what about if the problem was changed to ask for the sum of odd integers? can we amend the formula to get it? thanks!!
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 22 Oct 2010, 10:31
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even integers between 99 and 301 are 100, 102, 104.....,300

for an arithmatic progression,a1,a2,a3....,an the sum is n* (a1+an)/2

so here.. n = 101, hence sum = 101 * 400/2 = 101 * 200 = 20200 : B
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 22 Oct 2010, 16:03
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metallicafan wrote:
For any positive integer n, the sum of the first n positive integers equals \((n*(n+1))/2\). What is the sum of all the even integers between 99 and 301?
(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150


Sum = 100+102+...+300
= 100*101 + (0+2+4+...+200) [Taking 100 from each term in series, there are 101 terms]
= 100*101 + 2*(1+2+..+100) [Taking 2 common]
= 100*101 + 2*100*101/2 [Using formula given]
= 2*100*101
= 20200

Answer is (b)
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 07 Dec 2010, 07:30
can someone explain me why do we subtract 49 and no 50 ?
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 07 Dec 2010, 19:28
tt11234 wrote:
thanks!! it makes sense now...what about if the problem was changed to ask for the sum of odd integers? can we amend the formula to get it? thanks!!


The sum of n consecutive odd integers starting from 1 is \(n^2\)

e.g. sum of 1 + 3 + 5 + 7 = \(4^2\)

GMAT doesn't expect you to know it but it is a standard formula and doesn't hurt to know.
It can be easily derived using n(n+1)/2 formula. Let's wait for a day or two to see if someone comes up with a simple method of doing it.

tomchris wrote:
can someone explain me why do we subtract 49 and no 50 ?


We need to find the sum
50 + 51 + 52 +...+ 150 (The sum we need includes 50)

It can be done in two ways:

1. Using n(n + 1)/2

1 + 2 + 3 + ...49 + 50 + 51 + ....150 = 150.151/2
Also, 1 + 2 + 3 + ....+ 49 = 49.50/2

To get the required sum, we just subtract second equation from first.
We get: 50 + 51 + ...150 = 150.151/2 - 49.50/2

2. Using average of AP concept.
Note 50, 51...150 is an arithmetic progression. (Common difference between adjacent terms.)
The average of this AP = (first term + last term)/2 = (50 + 150)/2 = 100
Sum of AP = Average * No of terms = 100 * 101
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 08 Dec 2010, 18:32
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Ok. Let's quickly review sum of some series:

- Sum of positive consecutive integers starting from 1 is n(n + 1)/2 where n is the number of integers.
1+2+3+4+5+...+10 = 10*11/2

- Sum of positive consecutive even integers starting from 2 is n(n + 1) where n is the number of integers.
2+4+6+8+10 = 5*6 = 30

The explanation is simple.
2+4+6+8+10 = 2(1+2+3+4+5) By taking 2 common
Sum = 2(5*6/2) = 5*6

- Sum of positive odd integers starting from 1 is \(n^2\)
1+3+5+7+9+11 = \(6^2\)

I can derive it in the following way:
1+3+5+7+9+11 = (1+1)+(3+1)+(5+1)+(7+1)+(9+1)+(11 +1) - n
I add and subtract n from the right side.
The right side becomes: 2+4+6+8+10+12 - n = n(n+1) - n = \(n^2\)

So 100 + 102 + 104 + 106 + ...+ 300 can be solved in multiple ways.

It has been solved above in the following manner:

100 + 102 + 104 + 106 + ...+ 300 = 2( 50 + 51 + 52 + 53 + ...+ 150) Take out 2 common and find the sum in brackets.

50 + 51 + 52 + 53 + ...+ 150
We know the sum of consecutive integers but only when they start from 1. So what do we do?

We find the sum of first 150 numbers and subtract the sum of first 49 numbers from it. That will give us the sum of numbers from 50 to 150. Note that we subtract 49 numbers because 50 is part of our series. We need it so we do not subtract it.

50 + 51 + 52 + 53 + ...+ 150 = 150*151/2 - 49*50/2

Then 100 + 102 + 104 + 106 + ...+ 300 = 2( 150*151/2 - 49*50/2) = 20200
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 01 Jan 2011, 21:33
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There is an easier way (without using formula) :

number of even integers from 99 and 301 = (300-100)/2 + 1 = 101
Average = (100+300)/2 = 200

Sum of all even integers between 99 and 301 = 101 * 200 = 20,200
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 05 Feb 2011, 18:45
If there are 100 even integers in the sequence why do you +1 and multiply the average (200) by 101?
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 06 Feb 2011, 00:34
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kade wrote:
If there are 100 even integers in the sequence why do you +1 and multiply the average (200) by 101?


Question: even integers between 99 and 301

We know 99 is not even and 301 is not even; so ignore both of those

The actual sequence becomes:

100,102,104,106,108,.......,296,298,300

In the above sequence; the first number 100 is divisible by 2(or is even) and the last number 300 is also divisible by 2(or is even).

The common difference "d" is 2.

For such sequences; there is a formula to find the number of elements:

It is {(Last Number - First Number)/d} + 1

So; the above sequence actually contains {(300-200)/2}+1 = 101 elements

And Average = (first number + last number)/2 = (100+300)/2 = 200

Sum = Average * Number of Elements = 101 * 200 = 20200

Ans: 20200
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 17 Mar 2012, 16:45
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Since it is the even numbers, you can change the set to 100 to 300.

The average of the two is 200.

The amount of even integers is: ((300-100)/2)+ 1 = 101

200 * 101 = 20,200
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 03 Jun 2012, 17:53
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damham17 wrote:
For any positive integer n, the sum of the first n positive integers equals (n(n+1))/2. What is the sum of all the even integers between 99 and 301?

A) 10,100
B) 20,200
C) 22,650
D) 40,200
E) 45,150

Would someone please explain this? I do not quite understand the official explanation. Thanks in advance.



The sum of the first n terms of an arithmetic pogression is : n/2 *(2a + (n-1)d)

where a is the first term and d is the difference between consecutive terms.


we are to find out the sum of all even integers between 99 and 301

The even numbers look like 100,102,104....300

a=100
n=101
d=2

placing these values in the formula n/2 *(2a + (n-1)d)

sum of even integers = 101/2 * (2*100 + (101-1)*2)
=20200

Choice B.




Alternative method:


The sum of first n positive integers is n(n+1)/2

The sum of first 301 integers is 301*302/2=301*151=45451
The sum of first 99 integers is 99*100/2=99*50=4950

The sum of integers between 99 and 301 = 45451-4950=40501

The above contains all odd and even integers between 99 and 301 --- odd + even = 40501 ------->eqn 1



consider the following series:

numbers in series number of odd numbers sum of odd numbers(so) sum of even numbers(se) so-se
123 2 4 2 2
12345 3 9 6 3
1234567 4 16 12 4


You can see that in the first n integers, the difference of the sum of the odd numbers - sum of even numbers = number of odd numbers in the set.

between 1 and 301 we have 151 odd numbers
between 1 and 99 we have 50 odd numbers

difference of sum of odd numbers and sum of even numbers between 99 and 301 = 151-50=101 ----------> eqn 2

consider eqn 1 and eqn 2

odd+even=40501
odd-even=101

subtract 2 from 1

2even=40400
sum of even numbers between 99 and 301=20200

Choice B

Approach 3

you know that sum of odd + even between 99 and 301 is 40501 (same from previous approach)

The answer should be roughly half of this number --- 20250
closest is B.


Would love to see a simpler more elegant solution. It will save time for us.
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 04 Jun 2012, 02:21
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the sum of 1st even numbers = n(n+1)
the sum of first even numbers till 301 is 150*(150+1) =22650
the sum of first even numbers till 99 is 49*(49+1) =2450

22650-2450=20200

this method is not perfect. I wrote it just to show one more method
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 26 Aug 2012, 10:30
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In this question, they give you a formula to reference: the sum of the first n positive integers is n(n+1)/2

But wait, the question is about the sum of EVEN integers, not all positive integers.

We can understand the sum of even integers between 99 and 301 by somehow rewriting in terms of the expression they gave us. They were using the terminology of the first n positive integers.

OK, well sum of even integers between 99 and 301 is the same as between 100 and 300 - inclusive.

Same as the sum of the even integers up to 300 - sum of the even integers up to 98
Same as the sum of the first 150 even integers - sum of the first 49 even integers

So how do we translate this to the expression we were given? We need to translate our "even" integer expression into "positive integers" expression with our formula.

Sum of first 150 even integers = Sum of first 150 POSITIVE integers * 2
1, 2, 3, 4, 5 ===> 2, 4, 6, 8, 10 etc

Sum of first 49 even integers = Sum of first 49 POSITIVE integers * 2

2 * (n(n+1)/2) - 2 * (n(n+1)/2)

Where n=150 in the first case, and n = 49 in the second case

= 2* (150(151)/2) - 2 * (49(50)/2)
= 150*151 - 49*50
= 50 (3*151 - 49)
= 50 (453 - 49)
= 50 (404)
= 50 (400 + 4)
= 20,000 + 200
= 20,200
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 12 Feb 2013, 02:36
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Easier way would be using sum of all n even integers = n * (n+1)

Hence from 1-301 there are 150 even integers hence sum = 150 * 151 = 22650
From 1- 99 there are 49 even integers ( - 1 as 100 is not part of this) = 49 * 50 = 2450

So sum from 99-301 = 22650 -2450 = 20200
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Re: For any positive integer n, the sum of the first n positive integers   [#permalink] 12 Feb 2013, 02:36

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