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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.

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.

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?

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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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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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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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.

Although a formula is provided in this problem, we can easily solve it using a different formula:

sum = (average)(quantity)

Let’s first determine the average.

In any set of numbers in an arithmetic sequence, we can determine the average using the formula:

(1st number in set + last number in set)/2

Remember, we must average the first even integer in the set and the last even integer in the set. So we have:

(100 + 300)/2 = 400/2 = 200

Next we have to determine the quantity. Once again, we include the first even integer in the set and the last even integer in the set. Thus, we are actually determining the quantity of even consecutive even integers from 100 to 300, inclusive.

Two key points to recognize:

1) Because we are determining the number of “even integers” in the set, we must divide by 2 after subtracting our quantities.

2) Because we are counting the consecutive even integers from 100 to 300, inclusive, we must “add 1” after doing the subtraction.

quantity = (300 – 100)/2 + 1

quantity = 200/2 + 1

quantity = 101

Finally, we can determine the sum.

sum = 200 x 101

sum = 20,200

Note that the reason this is easier than using the formula provided, is that the given formula would have to be applied several times since we don’t want the total of all of the first 300 numbers. We’d have to remember to subtract the sum of the first 99 and divide by two to count only the even numbers. But we’d also have to account for the fact that the first and the last number in the set are both even. So even though a formula is given, it isn’t very easy to use.

Answer is B.
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For any positive integer n, the sum of the first n positive [#permalink]

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27 Jul 2017, 06:31

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Bang2919 wrote:

fameatop wrote:

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

I am not able to get the right answer using this formula. To find n: (300-100)/2 + 1 = 101. According to the formula mentioned it is n*(n+1) = 101 * 102. I am not understanding where i am going wrong

You are missing a very important point here. The question has mentioned that the sum of FIRST N POSITIVE INTEGERS EQUALS n(n+1)/2.

We have to just find the sum of even nos. from 100 to 300.

100,102,104,106,.....................,300

You have found correctly that the number of terms is 101.

But this formula can only be used for first n positive integers i.e. 1,2,3,4,5................................and so on.

Here the series is not starting from 1 but from 100.

Therefore

To use this formula, we will have to first find the sum of first 150 terms and then subtract the sum of first 49 terms.

Why so?

We have to find: 100+102+104+106+......................+300

This can also be written as: 2(50+51+.................+150)

We need the sum of 50th term+51st term, and so on till 150th term.

After you subtract the sum of 1st term till 49th term from the sum of 1st term till 150th term, multiply the answer by 2.

I hope you get it.

Otherwise there are some other great methods discussed here.

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

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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:

lll ask a stupid question 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 [#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?

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 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