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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]

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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
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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]

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New post 11 May 2010, 20:26
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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]

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New post 17 May 2010, 15:21
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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For any positive integer n, the sum of the first n positive integers [#permalink]

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New post 22 Oct 2010, 10:22
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
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

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even integers between 99 and 301 are 100, 102, 104.....,300

for an arithmatic progression,a1,a2,a3....,an the sum is n* (a1+an)/2

so here.. n = 101, hence sum = 101 * 400/2 = 101 * 200 = 20200 : B
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

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New post 22 Oct 2010, 16:03
metallicafan 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?
(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150


Sum = 100+102+...+300
= 100*101 + (0+2+4+...+200) [Taking 100 from each term in series, there are 101 terms]
= 100*101 + 2*(1+2+..+100) [Taking 2 common]
= 100*101 + 2*100*101/2 [Using formula given]
= 2*100*101
= 20200

Answer is (b)
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

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New post 24 Oct 2010, 11:21
We need to calculate 100 + 102 + 104 + 106 + ........+ 298 + 300 using the given formulae

This is = 2 [ 50 + 51 + 52 + ......+ 149 + 150]

= 2 [(1+ 2+3 + 4+5 + ......+ 149 + 150) - (1+2+3+.......+48 + 49)]

= 2[ 150*151/2 - 49*50/2 ]

= 50 [ 3*151 - 49 ] = 20200 (B)

There is no need to use any other formulae as N(a1+aN)/2 to solve this
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

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New post 01 Jan 2011, 20:12
#157: 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

Big kudos to someone who can get this right and explain the answer for me! :-D
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

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There is an easier way (without using formula) :

number of even integers from 99 and 301 = (300-100)/2 + 1 = 101
Average = (100+300)/2 = 200

Sum of all even integers between 99 and 301 = 101 * 200 = 20,200
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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

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New post 02 Jan 2011, 00:40
They way I would do it, using the given formula:

our sum = 100 + 102 + 104 + ... + 298 + 300 = 2*50 + 2*51 + 2*52 + .... + 2*149 + 2*150.
This is equal to: 2*(50 + 51 + 52 + .... + 149 + 150)= 2*(50 + 50+1 + 50+2 + .... + 50+99 + 50+100). Because in the parenthesis 50 appears 101 times, our sum can be written: 2*( 50*101 + 1 + 2 + .... + 99 + 100).

We can compute 1 + 2 +...+ 100 using the given formula, and 1 + 2 +.. + 100 = (100*101)/2=50*101.

So, the required sum=2*(50*101 + 50*101)=4*50*101=200*101=20.200
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Re: OG #157 [#permalink]

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New post 05 Feb 2011, 18:45
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]

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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]

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New post 16 Nov 2011, 19:50
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]

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Re: For any positive integer n, the sum of the first n positive [#permalink]

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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

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 integers [#permalink]

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Re: For any positive integer n, the sum of the first n positive integers   [#permalink] 24 Apr 2016, 13:03
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