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easier way to do simple question?

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by coolgirl26 » Wed Mar 04, 2009 6:09 pm
Hi all,

what is an easier way to solve this question

It took me about 5 minutes to solve because I had to list everything out.

So number 1) how can i solve this in an easier way? is there a formula

2) if the list is non inclusive of two numbers how can I also solve this in a simple fashion?

Thank you so much
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Source: — Problem Solving |

by Feep » Wed Mar 04, 2009 6:52 pm
The sum of any regularly spaced (i.e. arithmetic) sequence is quite simple: (a1 + an) * (n/2), where n is the number of elements in the sequence. Here, there are eleven elements (40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, this also happens to be the value of y), so let's plug in: (40 + 60) * (11/2) = 100 * 5.5 = 550 = x. x + y = 550 + 11 = 561.

This formula is markedly useful, so it's worth a quick memorization.
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by coolgirl26 » Wed Mar 04, 2009 7:41 pm
wow this helps alot!!!

thank you
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by mkbigmoz » Wed Mar 04, 2009 10:46 pm
Feep wrote:The sum of any regularly spaced (i.e. arithmetic) sequence is quite simple: (a1 + an) * (n/2), where n is the number of elements in the sequence. Here, there are eleven elements (40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, this also happens to be the value of y), so let's plug in: (40 + 60) * (11/2) = 100 * 5.5 = 550 = x. x + y = 550 + 11 = 561.

This formula is markedly useful, so it's worth a quick memorization.

You have no idea how long it would take me to complete these problems. I have purchased all the Kaplan/Manhattan books and I haven't come across one formula that addresses these types of problems. I would literally sit there and add up each sum, wasting 3 minutes for a ridiculous question.


I can not thank you enough.
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