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

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For any positive integer n, the sum of the first n positive [#permalink] New post 11 May 2010, 14:57
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For any positive integer n, the sum of the first n positive integers 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

OPEN DISCUSSION OF THIS QUESTION IS HERE: for-any-positive-integer-n-the-sum-of-the-first-n-positive-127817.html
[Reveal] Spoiler: OA
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Re: OG #157 [#permalink] New post 11 May 2010, 15:54
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CharmWithSubstance wrote:
For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

Answer is B: 20,200.

Don't understand the explanation - somehow they applied the above formula and customized it to even integers but I don't really follow. Can anyone explain the best way to solve? Thanks!


In any set of consecutive integers the sum = average*number of integers

average = 200
total = 150-50 + 1 = 101

150 because 300 is the 150th even integer
50 because 100 is the 50th even integer

101*200 = 20200
Therefore there are 101 integers
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Re: OG #157 [#permalink] New post 11 May 2010, 19:26
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CharmWithSubstance wrote:
For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

Answer is B: 20,200.

Don't understand the explanation - somehow they applied the above formula and customized it to even integers but I don't really follow. Can anyone explain the best way to solve? Thanks!

sum of all even integer 1 to 301 = 2 * (1+2+...+150) = (2*150*151)/2 = 150*151
sum of all even integer 1 to 99 = 2 * (1+2+....+49) = (2*49*50)/2 = 49*50

required sum = 150*151 - 49*50 = 50*(453 - 49) =
404 * 50 = 20200
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Re: OG #157 [#permalink] New post 17 May 2010, 14:21
u can also use the Aithmatic progression formula sum= n/2[2a + (n-1)d] here n=101 a=100 d=2 , put the values and you will get the answer as 20,200
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Re: OG #157 [#permalink] New post 05 Feb 2011, 17:45
If there are 100 even integers in the sequence why do you +1 and multiply the average (200) by 101?
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Re: OG #157 [#permalink] New post 05 Feb 2011, 23:34
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kade wrote:
If there are 100 even integers in the sequence why do you +1 and multiply the average (200) by 101?


Question: even integers between 99 and 301

We know 99 is not even and 301 is not even; so ignore both of those

The actual sequence becomes:

100,102,104,106,108,.......,296,298,300

In the above sequence; the first number 100 is divisible by 2(or is even) and the last number 300 is also divisible by 2(or is even).

The common difference "d" is 2.

For such sequences; there is a formula to find the number of elements:

It is {(Last Number - First Number)/d} + 1

So; the above sequence actually contains {(300-200)/2}+1 = 101 elements

And Average = (first number + last number)/2 = (100+300)/2 = 200

Sum = Average * Number of Elements = 101 * 200 = 20200

Ans: 20200
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Re: OG #157 [#permalink] New post 16 Nov 2011, 18:50
fluke wrote:
kade wrote:
If there are 100 even integers in the sequence why do you +1 and multiply the average (200) by 101?


Question: even integers between 99 and 301

We know 99 is not even and 301 is not even; so ignore both of those

The actual sequence becomes:

100,102,104,106,108,.......,296,298,300

In the above sequence; the first number 100 is divisible by 2(or is even) and the last number 300 is also divisible by 2(or is even).

The common difference "d" is 2.

For such sequences; there is a formula to find the number of elements:

It is {(Last Number - First Number)/d} + 1

So; the above sequence actually contains {(300-200)/2}+1 = 101 elements

And Average = (first number + last number)/2 = (100+300)/2 = 200

Sum = Average * Number of Elements = 101 * 200 = 20200

Ans: 20200


Your answer is correct, but the red text above contains an error. The last number is 300 and the first number is 100 therefore there are 101 elements. I think your work above should read {(300-100)/2}+1 = 101
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Re: OG #157 [#permalink] New post 28 Oct 2013, 06:47
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Re: For any positive integer n, the sum of the first n positive [#permalink] New post 28 Oct 2013, 07:09
Expert's post
For any positive integer n, the sum of the first n positive integers 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

Approach #1:
Even integer between 99 and 301 represent evenly spaced set (aka arithmetic progression): 100, 102, 104, ..., 300. Now, the sum of the elements in any evenly spaced set is the mean (average) multiplied by the number of terms. (Check Number Theory chapter of Math Book for more: math-number-theory-88376.html)

Average of the set: (largest+smallest)/2=(300+100)/2=200;
# of terms: (largest-smallest)/2+1=(300-100)/2+1=101 (check this: totally-basic-94862.html#p730075);

The sum = 200*101= 20,200.

Answer: B.

Approach #2:
Using the formula of the sum of the first n positive integers: n(n+1)/2.

100+102+...+300=2(50+51+..+150). Now, the sum of the integers from 50 to 150, inclusive equals to the sum of the integers from 1 to 150, inclusive minus the sum of the integers from 1 to 49, inclusive. 2(50+51+..+150)=2*(150(150+1)/2-49(49+1)/2)=20,200.

Answer: B.

OPEN DISCUSSION OF THIS QUESTION IS HERE: for-any-positive-integer-n-the-sum-of-the-first-n-positive-127817.html
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Re: For any positive integer n, the sum of the first n positive   [#permalink] 28 Oct 2013, 07:09
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