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

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

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19 Feb 2012, 19:19
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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

Hi,

Hoping you can explain this one to me.
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Re: For any positive integer n, the sum.... [#permalink]

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

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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: 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: totally-basic-94862.html#p730075);

The sum = 200*101= 20,200.

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.

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

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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: What is the sum of all even integers between 99 and 301? [#permalink]

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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 [#permalink]

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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 [#permalink]

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16 Jul 2012, 12:19
but whats the purpose of the formula in question itself?
Quote:
the sum of the first n positive integers equals n(n+1)/2

if there is no need in it?
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For any positive integer n, the sum of the first n positive [#permalink]

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19 Aug 2012, 13:01
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This problem can be solved with an alternative formula:-
Sum of first 'n' even integers is given by - n(n+1), where n=2,4,6,......any even integer
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Re: For any positive integer n, the sum of the first n positive [#permalink]

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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 [#permalink]

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01 Sep 2012, 06:26
Galiya wrote:
but whats the purpose of the formula in question itself?
Quote:
the sum of the first n positive integers equals n(n+1)/2

if there is no need in it?

The way I see it the formula is there just to make the question a bit harder. An evil trick, if you want.

The OG explanation is equally confusing and I personally do not think is worth the effort to try and master it, there are much "cleaner" ways to solve this problem, as many users here have shown.
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For any positive integer n, the sum of the first n positive inte [#permalink]

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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 [#permalink]

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25 Feb 2013, 08:42
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Hi neerajeai, please note that the n(n+1)/2 shortcut formula is only applicable if the starting point is 1. Anytime you want to find the number of terms between two given numbers you should use the general formula ((first - last) / frequency) + 1. You can also multiply by the average at the end to get the sum.

In your case it is (((301-99)/2) + 1) * 200 = 102 * 200 = 20400. Answer choice C with a presumed typo in the unit digit?

Hope this helps!
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What is the sum of all even integers between 99 and 301? [#permalink]

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18 Nov 2014, 07:42
What is the sum of all even integers between 99 and 301?
a. 10,100
b. 20,200
c. 22,650
d. 40,200
e. 45,150
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Re: For any positive integer n, the sum of the first n positive [#permalink]

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18 Nov 2014, 07:46
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Blackbox wrote:
What is the sum of all even integers between 99 and 301?
a. 10,100
b. 20,200
c. 22,650
d. 40,200
e. 45,150

Merging topics. Please search before posting.
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For any positive integer n, the sum of the first n positive [#permalink]

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18 Nov 2014, 08:13
If I were to use this formula for sum of numbers in an AP ---> $$\frac{n}{2}*(2a + (n-1)d)$$, I'd first find the number of terms in the collection using
n=$$\frac{(Last Term - First Term)}{2} - 1$$ (- 1 because I am excluding the first and last terms)

So, that would give n=$$\frac{(301-99)}{2}-1$$ = 100
Therefore,
a=99
n=100
d=2
Replacing these values in the formula n/2 *(2a + (n-1)d), we get sum of even integers =$$\frac{100}{2} * (2*99+ (100-1)*2)$$ = 19,800. Where am I wrong?
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For any positive integer n, the sum of the first n positive [#permalink]

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18 Nov 2014, 08:34
OK, I think I now understand where I went wrong. The question asks- Find the sum of multiples of 2. So, the first multiple of 2 in the collection begins at 100 and last one ends at 300. Therefore, your first term is 100 and last 300. The formula for n changes though:

n=$$\frac{(Last Term - First Term)}{2} + 1$$, which yields 101.

The value of a also changes because the first term is 100 and NOT 99. Sub-in those values,
$$\frac{101}{2} * (2*100+ (101-1)*2)$$ = 20,200

For reference, please see Bunuel's post here >>http://gmatclub.com/forum/how-many-multiples-of-4-are-there-between-12-and-94862.html#p730075

My question still is, how would the formula for n change if the question asks:
1.Including both the terms
2. Including at least one term.
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For any positive integer n, the sum of the first n positive [#permalink]

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18 Nov 2014, 08:55
Blackbox wrote:
If I were to use this formula for sum of numbers in an AP ---> $$\frac{n}{2}*(2a + (n-1)d)$$, I'd first find the number of terms in the collection using
n=$$\frac{(Last Term - First Term)}{2} - 1$$ (- 1 because I am excluding the first and last terms)

So, that would give n=$$\frac{(301-99)}{2}-1$$ = 100
Therefore,
a=99
n=100
d=2
Replacing these values in the formula n/2 *(2a + (n-1)d), we get sum of even integers =$$\frac{100}{2} * (2*99+ (100-1)*2)$$ = 19,800. Where am I wrong?

the formula, you are using for calculating the total number of terms is wrong. because you shouldn't put '-1' at the end of expression. why ??
consider this. for any arithmetic sequence, nth term is given as tn= a+(n-1)d. now here, if we consider a=99 and tn+1=301. and d=1, that is all terms are consecutive, then we have n-1=301-99, n=203. i.e we have total of 203 terms in the expression. if i remove 99 from the sequence. ( which you were trying to do in a wrong way.). then sequence will begin from 100 and ends at 301. i.e we have total of 202 terms. now of these terms. half are even, and half are odd. i.e. we have 101 even terms and 101 odd terms

now put the same thing in your formula of n/2 *(2a + (n-1)d), n=101 and a=100 (why ?? because we have already discarded 99) and d=2
101/2(200+200)=20,200
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Re: For any positive integer n, the sum of the first n positive [#permalink]

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

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23 Feb 2016, 03:27
Galiya wrote:
but whats the purpose of the formula in question itself?
if there is no need in it?

Well I used a quick and dirty way to solve it using the formula. Use the formula to add ALL integers from 1 to 300 inclusive, then subtract all the numbers from 1 to 99, also using the formula, and you get the sum of all integers from 100 to 300 inclusive: 40,200.
Then understand that once you remove all the odd integers, the answer will be almost exactly half of 40,200. The only close option was 20,200 (B).
For any positive integer n, the sum of the first n positive   [#permalink] 23 Feb 2016, 03:27

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