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Consecutive Integers

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by bfman » Tue Jul 21, 2009 11:30 am
Official Guide Ed 12. Page 236. Q 157.

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?

When I solved it, I ignored the first [n(n+1)]/2 part.

I used the MGMAT method of getting sum of a sequence of even numbers:

Total: 300-100+1 = 201/2 = 100.5
Average: 300+100/2 = 200

100.5 * 200 = 20,100. Very close to the correct answer which is 20,200.
What am I doing wrong here?

The OG does use the provided algebraic expression, but I am wondering if this can be solved without it.
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by jjk » Tue Jul 21, 2009 11:39 am
You're actually supposed to find the sum of the first and last terms of the series, divide the result by 2, and then multiply that by the amount of terms in the series.

100 and 300 are the first and last even positive integers in that series.

100 + 300 = 400

Next, divide the sum (400) by 2.

400/2 = 200

Multiply this by the amount of terms in the series. From 100 to 300, inclusive, there are 101 positive even integers.

200 X 101 = 20,200
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Re: Consecutive Integers

by Stuart@KaplanGMAT » Tue Jul 21, 2009 11:44 am
bfman wrote:
I used the MGMAT method of getting sum of a sequence of even numbers:

Total: 300-100+1 = 201/2 = 100.5
The formula is:

(last number - first number)/2 + 1

So:

(300 - 100)/2 + 1 = 200/2 + 1 = 101

You added the 1 before dividing, which is why your answer was off.
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by bfman » Tue Jul 21, 2009 12:48 pm
Thank you!
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by drabblejhu » Wed Sep 16, 2009 9:10 pm
I also did it the MGMAT strategy guide way of
(# terms) ( first plus last term)/2. Anyone have a sense of whether we'll ever ONLY be able to use the summation method given? It just tripped me up. Hoping to be able to avoid. Thanks!
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by bpgen » Sat Mar 06, 2010 8:39 pm
Stuart Kovinsky wrote:
bfman wrote:
I used the MGMAT method of getting sum of a sequence of even numbers:

Total: 300-100+1 = 201/2 = 100.5
The formula is:

(last number - first number)/2 + 1

So:

(300 - 100)/2 + 1 = 200/2 + 1 = 101

You added the 1 before dividing, which is why your answer was off.
Hi Stuart, I disagree..see my explanation below. Correct me if I'm wrong.

Let for e.g, sequence are 100,101,102,103, how many even integers are there?
(103-100+1)/2=2, Correct.
using your formula it would be (103-100)/2+1=2.5~3, Not Correct.
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by Vorskl » Sat Oct 16, 2010 8:14 am
Hi Stuart, I disagree..see my explanation below. Correct me if I'm wrong.

Let for e.g, sequence are 100,101,102,103, how many even integers are there?
(103-100+1)/2=2, Correct.
using your formula it would be (103-100)/2+1=2.5~3, Not Correct
Man, the formula is correct. If you want to know the number of even numbers in a serie, your serie shall start and finish with an even number: 100 - 102 (or 104, does not matter). Let's recalculate:

(102-100)/2+1 = 1+1 = 2
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