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Average - Even integers

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Average - Even integers

by Xbond » Thu Jun 04, 2009 9:48 pm
Hi there,

Could you explain to me the concept. I didn't succeed to find the solution right away


The sum of the first 50 positive integers is 2550. What is the seum of the even integers from 102 to 200, inclusive ?

A) 5100 B) 7550 C) 10 100 D) 15 500 E) 20 100


Answser B
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by rah_pandey » Thu Jun 04, 2009 11:08 pm
I think the concept to use is

given: 1+2+3+.....50=2550 (It should be 1275)

to find 102+104+....200

take 2 common from each
=2*(51+52+53+....100)
now 51=50+1
52=50+2
53=50+3
100=50+50
=>2*(50*50+1+2+3+......50)
=>2*(2500+1275)=7550

basically why sum of first 50 integers is given is to solve question without knowing formula for sum of first n integers
you can also find the answer by
2*(51+52+....100)=2*(sumof 1st 100-sum of first 50)
sum of first N integers is

(N*N+1)/2

=2*(100*101/2-1275)=100*101-2550=10100-2500=7550


I hope i am clear
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by joyseychow » Fri Jun 05, 2009 2:17 am
rah_pandey wrote:I think the concept to use is

given: 1+2+3+.....50=2550 (It should be 1275)

to find 102+104+....200

take 2 common from each
=2*(51+52+53+....100)
now 51=50+1
52=50+2
53=50+3
100=50+50
=>2*(50*50+1+2+3+......50)
=>2*(2500+1275)=7550

basically why sum of first 50 integers is given is to solve question without knowing formula for sum of first n integers
you can also find the answer by
2*(51+52+....100)=2*(sumof 1st 100-sum of first 50)
sum of first N integers is

(N*N+1)/2

=2*(100*101/2-1275)=100*101-2550=10100-2500=7550


I hope i am clear
I don't quite understand your explanation. What do you mean by
"given: 1+2+3+.....50=2550 (It should be 1275)"?


I got the answer without using the sum of first 50 integers.

Average:
1st even term + Last even term (between 102 and 200 inclusive)/2
-----> (102+200)/2=151

Term:
Last even term - 1st even term (between 102 and 200 inclusive)/ 2 +1
-----> (200-102)/2 + 1=50

Hence, Sum=average/term
-----> 151*50=7550
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by rah_pandey » Fri Jun 05, 2009 2:38 am
Ok,
I don't quite understand your explanation. What do you mean by
"given: 1+2+3+.....50=2550 (It should be 1275)"?
I thought this info was given as a part of question
The sum of the first 50 positive integers is 2550.
This is wrong. sum of first 50 positive int is 1275

In objective question if your answer is right and method makes sense than there should be no problem

Is the formula you have used a standard one. It will be good to know its limitation(when you should not apply it) but for this question it solves the purpose. I was trying to use the info given and than solve the qn.
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by ghacker » Fri Jun 05, 2009 10:13 am
If we take the sum of the first positive integers as X , what ever the value

we definitely know the value of the sum of even numbers from 102 to 200 inclusive . How ?

100+2 =102
100 +4 = 104
............................
...........................
100+100 = 200

so when we add we get = 100*50+Sum(2 to 100)
If we take 2 as the common factor from Sum(2 to 100)
we have 2(1+2+3+.........+50)= 2*X

Total = 5000+2*X

I think you wanted to say the sum of 1st 50 even integers
2X = 2550

Hence the sum =5000+2550= 7550
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by Xbond » Mon Jun 08, 2009 1:19 pm
Many thks, Joyseyshow
your explanation is very clear.
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by Xbond » Mon Jun 08, 2009 1:25 pm
By the way, do you really think that is useful to get data of the sum of the first 50 positive integers is 2550 ?
because as youknow, we can resolve it without it, isn't it ?
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by Stuart@KaplanGMAT » Mon Jun 08, 2009 1:27 pm
First important thing - there's a typo in the question.

The question should read "the sum of the first 50 EVEN positive integers is 2550..."

While you don't need that information to answer the question (applying the average of a set of consecutive numbers formula works great), it can be used to quickly sum the next 50 even positive integers, {102, ... 200}.

We can see that each number in the set {2, 4, ... , 100} is exactly 100 less than a number in the set {102, 104, ..., 200}.

Therefore the sum of the set {102, 104, ..., 200} will be 100*(# of terms) more than the sum of the set {2, 4, ..., 100}.

We're told that there are 50 terms, so the answer to the question is:

2550 + 100(50) = 2550 + 5000 = 7550
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by Xbond » Mon Jun 08, 2009 1:42 pm
How do you calculate quickly the first 50 EVEN positive integers ? sorry, this evening, I burn out. I have some difficulties to think.
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by Stuart@KaplanGMAT » Mon Jun 08, 2009 3:01 pm
Xbond wrote:How do you calculate quickly the first 50 EVEN positive integers ? sorry, this evening, I burn out. I have some difficulties to think.
Average = (Sum of terms)/(# of terms)

so

Sum of terms = Average * (# of terms)

To find the average of a set of consecutive numbers, we simply take:

(smallest + biggest)/2

So, for the first 50 even positive numbers, the average is:

(2 + 100)/2 = 102/2 = 51

and the sum is:

51 * 50 = 2550
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by Xbond » Mon Jun 08, 2009 8:16 pm
many thks
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by joyseychow » Sat Jun 13, 2009 7:34 am
Stuart, thanks so much for clarifying!! :D

Xbond, it is good to know both approach. It comes in handy.
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