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Re: For any positive integer n, the sum of the first n positive [#permalink]
19 Feb 2012, 22:11

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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.

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.

Re: For any positive integer n, the sum of the first n positive [#permalink]
04 Jun 2012, 01:21

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]
19 Aug 2012, 12:01

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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]
26 Aug 2012, 09:30

Expert's post

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

Re: For any positive integer n, the sum of the first n positive [#permalink]
01 Sep 2012, 05:26

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. _________________

For any positive integer n, the sum of the first n positive inte [#permalink]
12 Feb 2013, 01:36

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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 _________________

Re: For any positive integer n, the sum of the first n positive [#permalink]
25 Feb 2013, 07:42

1

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Expert's post

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?

Re: For any positive integer n, the sum of the first n positive [#permalink]
06 Nov 2014, 23:43

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For any positive integer n, the sum of the first n positive [#permalink]
18 Nov 2014, 07:13

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]
18 Nov 2014, 07:34

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]
18 Nov 2014, 07:55

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

gmatclubot

For any positive integer n, the sum of the first n positive
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18 Nov 2014, 07:55

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