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The sum of the first 50 positive even integers is 2,550.

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The sum of the first 50 positive even integers is 2,550. [#permalink] New post 27 Apr 2008, 18:33
The sum of the first 50 positive even integers is 2,550. What is the sum of the even integers from 102 to 200 inclusive.
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Re: AP series , need quick way to solve this [#permalink] New post 27 Apr 2008, 19:19
Even numbers from 102 to 200 = 2* ( 51 to 100 ).
No of elements in this series = (100-51+1) = 50.
The middle value of this series = (51+100) / 2 = 75.5
The sum = 2 * 75.5 * 50 = 7550.
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Re: AP series , need quick way to solve this [#permalink] New post 27 Apr 2008, 21:11
bsd_lover wrote:
Even numbers from 102 to 200 = 2* ( 51 to 100 ).
No of elements in this series = (100-51+1) = 50.
The middle value of this series = (51+100) / 2 = 75.5
The sum = 2 * 75.5 * 50 = 7550.


Not understanding this at all...

bsd can you elaborate on your explanation?
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Re: AP series , need quick way to solve this [#permalink] New post 27 Apr 2008, 21:18
Hi, this is not really "my" method. Its explained really well in the Manhattan GMAT series.

Say if are finding out the sum of a series lets use 1 - 99 for example.

First thing to find out is how many items in the series. In this case - there are 99 items.

Next thing to find out is what is the average of all the items - this is the absolute middle value. This can be found simply by taking the average of first and last elements : 99+1 / 2 = 50 <-- obviously the middle value.

So now we have an average (i.e. a mean) and a number of elements. To find the sum we simply multiply the mean with the number of elements : 50 * 99 = 4950

cheers.
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Re: AP series , need quick way to solve this [#permalink] New post 27 Apr 2008, 21:29
bsd_lover wrote:
Hi, this is not really "my" method. Its explained really well in the Manhattan GMAT series.

Say if are finding out the sum of a series lets use 1 - 99 for example.

First thing to find out is how many items in the series. In this case - there are 99 items.

Next thing to find out is what is the average of all the items - this is the absolute middle value. This can be found simply by taking the average of first and last elements : 99+1 / 2 = 50 <-- obviously the middle value.

So now we have an average (i.e. a mean) and a number of elements. To find the sum we simply multiply the mean with the number of elements : 50 * 99 = 4950

cheers.



brilliant! that cleared things up very nicely.

does this have anything to do w/ the question at hand: The sum of the first 50 positive even integers is 2,550.
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Re: AP series , need quick way to solve this [#permalink] New post 27 Apr 2008, 21:42
droopy57 wrote:
The sum of the first 50 positive even integers is 2,550


I also ask myself "why ETS adds this stuff to the text?" maybe it will be a lurring confusal! :lol:
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Re: AP series , need quick way to solve this [#permalink] New post 27 Apr 2008, 21:49
Red herring ?? I dunno ... We dont really need that info to solve the problem quickly.
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Re: AP series , need quick way to solve this [#permalink] New post 29 Apr 2008, 01:20
droopy57 wrote:
brilliant! that cleared things up very nicely.
does this have anything to do w/ the question at hand: The sum of the first 50 positive even integers is 2,550.

2,550 can be used for another way of solving:
the sum of first 50 positive inegers looks like 2 + 4 + 6 + .. + 100, all in all 50 items
the sum of next 50 positive integers looks like 2+100 + 4+100 + 6+100 + ... + 100+100, so you can find the sum of even integers from 102 to 200 by adding: 2550 + 100*50 = 7550

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Re: AP series , need quick way to solve this [#permalink] New post 29 Apr 2008, 09:51
Sunchaser20 wrote:
droopy57 wrote:
brilliant! that cleared things up very nicely.
does this have anything to do w/ the question at hand: The sum of the first 50 positive even integers is 2,550.

2,550 can be used for another way of solving:
the sum of first 50 positive inegers looks like 2 + 4 + 6 + .. + 100, all in all 50 items
the sum of next 50 positive integers looks like 2+100 + 4+100 + 6+100 + ... + 100+100, so you can find the sum of even integers from 102 to 200 by adding: 2550 + 100*50 = 7550

BR


Brilliant. I did the same way.
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Re: AP series , need quick way to solve this [#permalink] New post 29 Apr 2008, 19:19
bsd,

Why do you use multiplier of 2 in the problem? Yet in the example you provide there is no multiplier. I understand how you derive the 2 using property associations; however, if there are fifty elements in the problem why is the answer not 50*75.5?
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Re: AP series , need quick way to solve this [#permalink] New post 29 Apr 2008, 19:32
Its because the original question deals with a sequence of EVEN integers - which I convert to a regular sequence of integers by introducing the 2.

My second example is generic and uses a regular sequence.

smginnis wrote:
bsd,

Why do you use multiplier of 2 in the problem? Yet in the example you provide there is no multiplier. I understand how you derive the 2 using property associations; however, if there are fifty elements in the problem why is the answer not 50*75.5?
Re: AP series , need quick way to solve this   [#permalink] 29 Apr 2008, 19:32
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