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For any positive integer n, the sum of the first n position

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For any positive integer n, the sum of the first n position

by mitzwillrockgmat » Sun Jun 13, 2010 8:44 am
For any positive integer n, the sum of the first n positive integers equals [n(n+1)]/2. What is the sum of all even integers between 99 and 301?

A.10,100
B.20,200
C.22,650
D.40,200
E.45,150

Hello!

I have a strange issue here. Given the above equation; [n(n+1)]/2 I assumed that I HAD to apply this equation to get the answer. However, the answer can only be found by using the standard AP sum formula [N * (1st term + last term)]/2 . So what is the point of giving that above?! I found this unnecessarily misleading!

to make my point:

sum of even no. between 99 and 301.

between 99 & 301 there are 101 even no.s

so if i use formula above n(n+1)/2 = [101*102]/2 = 5151

if i use the correct formula 101/2 *100+300 = 20,200

Can someone please explain why would they do that???????
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by albatross86 » Sun Jun 13, 2010 9:14 am
Your approach is correct, and I guess they included the sum of first n numbers just for those people who don't know standard AP formula, so that they could derive it from there.

For example, since they have given you sum of first n integers, the sum of first n EVEN integers will therefore be twice that.

(Since Sum of first n integers = 1 + 2 +3 +...+ n and Sum of first n EVEN integers = 2 + 4 + 6 +....+ 2n)

So you have the formula as n(n+1)

You can then find the sum upto 98, which would be 49*50 = 2450
Then find the sum upto 300, which would be 150*151 = 22650

The sum from 99 to 301 would therefore be 22650 - 2450 = 20,200

This is done so that people who are unaware of the formula for a standard AP can still solve the problem :)
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by mitzwillrockgmat » Sun Jun 13, 2010 9:26 am
albatross86 wrote:Your approach is correct, and I guess they included the sum of first n numbers just for those people who don't know standard AP formula, so that they could derive it from there.

For example, since they have given you sum of first n integers, the sum of first n EVEN integers will therefore be twice that.

(Since Sum of first n integers = 1 + 2 +3 +...+ n and Sum of first n EVEN integers = 2 + 4 + 6 +....+ 2n)

So you have the formula as n(n+1)

You can then find the sum upto 98, which would be 49*50 = 2450
Then find the sum upto 300, which would be 150*151 = 22650

The sum from 99 to 301 would therefore be 22650 - 2450 = 20,200

This is done so that people who are unaware of the formula for a standard AP can still solve the problem :)
But the formula given is [n*n +1]/2 so shouldn't it be 49*50/2 =1225 ?
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by albatross86 » Sun Jun 13, 2010 9:28 am
n(n+1) /2 is for the sum of first n integers. ( 1 + 2 + 3+ .... + n)

The sum of first n EVEN integers is TWICE that. ( 2 + 4 + 6 +... + 2n) = 2 * ( 1+ 2 + 3 +...+n) = 2* n(n+1)/2 = n*(n+1)

Make sense? :)
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by mitzwillrockgmat » Sun Jun 13, 2010 9:36 am
albatross86 wrote:n(n+1) /2 is for the sum of first n integers. ( 1 + 2 + 3+ .... + n)

The sum of first n EVEN integers is TWICE that. ( 2 + 4 + 6 +... + 2n) = 2 * ( 1+ 2 + 3 +...+n) = 2* n(n+1)/2 = n*(n+1)

Make sense? :)
Hmmm well it makes sense NOW but i still find this annoying...anyone who's giving the gmat must've learned the standard formula so giving it like this is just weird & confusing! i was doing this question on a practice test so because of this i wasted 10 minutes!

to be honest, i really wouldn't figure out during the exam that i should see the given equation as 2* n(n+1)/2 = n*(n+1)

So, I guess, the lesson from this question is...if you know a question involves a standard formula just stick to that rather than the one given.

Thanks anyways!
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by odod » Sun Jun 13, 2010 2:29 pm
hello - I'm not sure what the 'standard' formula' that the first post describes. Can anyone help me out explaining it here?

thanks
ODOD
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by mitzwillrockgmat » Sun Jun 13, 2010 3:39 pm
odod wrote:hello - I'm not sure what the 'standard' formula' that the first post describes. Can anyone help me out explaining it here?

thanks
hi,

its n/2 *1st +last term ..........n=no. of terms in the series

so for this series there are 101 terms, the 1st term is 100 & the last term is 300.

so,

101/2 *100+300 = 20,200
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by sogmat » Sun Jun 13, 2010 4:36 pm
How do you get n=101?
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by mitzwillrockgmat » Sun Jun 13, 2010 9:26 pm
sogmat wrote:How do you get n=101?
the nos are from 100 to 300

between 100 & 300 there are 300-100+1 = 201 nos

in a series of consecutive nos half the no.s will be even (i.e. check 1 to 10)

so given that the end points are both even, there are 101 even nos
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by rbansal » Sun May 01, 2011 7:51 pm
Is the standard formula only used for the sum of consecutive integers?

RB
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by GMATGuruNY » Mon May 02, 2011 3:03 am
For any positive integer n, the sum of first n positive integer equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A) 10,100
B) 20,200
C) 22,650
D) 40,200
E) 45,150
The question asks for the sum of the even integers from 100 to 300, inclusive.

Ignore the formula given. Instead, the following can be used to calculate the sum of any set of evenly spaced integers:

Sum = (number of integers) * (average of biggest and smallest)

To count the number of evenly spaced integers in a set:

Number of integers = (biggest - smallest)/interval + 1

The INTERVAL is the distance between one term and the next.
Since we're adding only the even integers here, the interval is 2.
Thus, the number of even integers from 100 to 300 = (300-100)/2 + 1 = 101.
Average of biggest and smallest = (300+100)/2 = 200.
Sum = (number of integers) * (average of biggest and smallest) = 101*200 = 20,200.

The correct answer is B.
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by rbansal » Mon May 02, 2011 8:23 am
GMATGuruNY wrote:
For any positive integer n, the sum of first n positive integer equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A) 10,100
B) 20,200
C) 22,650
D) 40,200
E) 45,150
The question asks for the sum of the even integers from 100 to 300, inclusive.

Ignore the formula given. Instead, the following can be used to calculate the sum of any set of evenly spaced integers:

Sum = (number of integers) * (average of biggest and smallest)

To count the number of evenly spaced integers in a set:

Number of integers = (Biggest - Smallest)/(distance between each successive pair) + 1

Since we're adding only the even integers, the distance between each successive pair is 2.
Thus, the number of even integers from 100 to 300 = (300-100)/2 + 1 = 101.
Average of biggest and smallest = (300+100)/2 = 200.
Sum = (number of integers) * (average of biggest and smallest) = 101*200 = 20,200.

The correct answer is B.
So we just divide by whatever the increment is, correct?
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by fueledGMAT » Wed Nov 30, 2011 10:43 am
I have to agree with the OP that the formula given on this problem was completely misleading. I even had the right answer but continued to plug into the formula given and kept ending up with A instead of B. This is the type of question I fear the most on the GMAT because it makes me second guess formulas I am already comfortable with.
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by Cally627 » Sat Apr 28, 2012 1:51 pm
Hi, maybe i've been studying for wayy to long today where i've lost my mind but i have a question regarding this problem.

I see how you solve the problem (i used the same formula)

sum = avg x number of integers
avg = largest + smallest / 2
number of integers = largest - smallest / (increment - in which case 2 - bc events) + 1

however.... to find the number of integers - i was using 301-99 / 2 + 1 = 102
*it says on page 51 of mgmat that you are to add the extra 1 to the equation.

i see that you guys are using 100, and 300 instead of the 99 and 301.... (is this bc it's asking for even integers)... a little lost.

if someone could clear this up - it would help. thanks!
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