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Re: AP series , need quick way to solve this [#permalink]

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27 Apr 2008, 20:19

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.

Re: AP series , need quick way to solve this [#permalink]

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27 Apr 2008, 22:11

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.

Re: AP series , need quick way to solve this [#permalink]

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27 Apr 2008, 22:18

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

Re: AP series , need quick way to solve this [#permalink]

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27 Apr 2008, 22:29

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.

Re: AP series , need quick way to solve this [#permalink]

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29 Apr 2008, 02:20

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

Re: AP series , need quick way to solve this [#permalink]

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29 Apr 2008, 10:51

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

Re: AP series , need quick way to solve this [#permalink]

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29 Apr 2008, 20:19

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?

Re: AP series , need quick way to solve this [#permalink]

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29 Apr 2008, 20:32

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?