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The sum of the first 50 positive even integers is 2550. What is the

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The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

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

A. 5,100
B. 7,550
C. 10,100
D. 15,500
E. 20,100

Originally posted by CasperMonday on 03 Sep 2009, 06:35.
Last edited by Bunuel on 11 Oct 2019, 05:29, edited 4 times in total.
Renamed the topic and edited the question.
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The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

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New post 21 Mar 2012, 13:45
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enigma123 wrote:
The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102 to 200 inclusive?

A. 5,100
B. 7,550
C. 10,100
D. 15,500
E. 20,100


# of even terms (multiples of 2) in the range from 102 to 200, inclusive is \(\frac{(last - first)}{2} + 1=\frac{(200-102)}{2}+1=50\);

\(Mean = \frac{(last+first)}{2}\);

\(The \ sum = (mean)*(number \ of \ terms) = \frac{(200+102)}{2}*50 = 7,550\).

Answer: B.
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Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

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New post 03 Sep 2009, 06:53
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CasperMonday wrote:
The sum of the first 50 positive even integers is 2550. What is the sum of even integers from 102 to 200 inclusive?
a. 5100
b. 7550
c. 10100
d. 15500
e. 20100

OA


I had derived a formula for calculating the summation of even/odd numbers in a series, when I was in high school.

Sum of EVEN numbers in a series from 1 -> n (where n is the LAST EVEN number in a series) = n*(n+2)/4
Sum of ODD numbers in a series from 1 -> n (where n is the LAST ODD number in a series) = (n+1)*(n+1)/4

This works perfect for any series starting from 1. Using the above formula we can solve the problem:

200*202/4 - 100*102/4 = 7550

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Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

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New post 03 Sep 2009, 06:44
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My solution is:

First 50 even integers:
2
4
6
8
<...>

Integers from 102 to 200
102
104
106
108
<...>

We notice that each integer from the second set is 100 more than the respective integer in the first set. Since we have 50 even integers from 102 to 200, then:
2550+(100*50)=7550.

Are there any other ways to solve the prob? Thanks!

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Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

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New post 03 Sep 2009, 06:58
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CasperMonday wrote:
The sum of the first 50 positive even integers is 2550. What is the sum of even integers from 102 to 200 inclusive?
a. 5100
b. 7550
c. 10100
d. 15500
e. 20100

OA


Sum of even integers from 102-200(both included)

No of terms = (200-102)/2 + 1 = 50 (+1 because both are included)

If you observe, this is AP (Arithmetic progression): 102,104,106,......,200
Sum of n terms in AP Formula, Sum = n/2[2a + (n-1)d]
where,
n = no of terms = 50 in this case.
a = starting number = 102
d = difference between AP = 2

So, Sum = 50/2[2*102 + (50-1)2]

Solving this, we can get the answer.

This is one more solution for this kind of problems.
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Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

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New post 03 Sep 2009, 08:23
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Sum of even integers from 102-200(both included)

No of terms = (200-102)/2 + 1 = 50 (+1 because both are included)

If you observe, this is AP (Arithmetic progression): 102,104,106,......,200
Sum of n terms in AP Formula, Sum = n/2[2a + (n-1)d]
where,
n = no of terms = 50 in this case.
a = starting number = 102
d = difference between AP = 2

So, Sum = 50/2[2*102 + (50-1)2]

Solving this, we can get the answer.

This is one more solution for this kind of problems.

Saurabhricha,
what would be the formula for odd nbrs?
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Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

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New post 21 Mar 2012, 14:01
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Sum = Avg * number of items

Avg = (200 + 102) / 2 = 302 / 2 = 151

No. of items = 200-102 = 98/2 (Only the even numbers so divide by 2) = 49 + 1 (add 1 the last item) = 50

Sum = 151 * 50 = 7550

ANS : B
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Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

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New post Updated on: 16 Jun 2015, 22:01
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Even easier method:

First 50 even integers are 2, 4, 6, ... , 100. Sum is stated as 2550.
Sum of even integers between 102 and 200 are 100 more than each integer in the previous sum (102 is 100 more than 2, etc.).

So...when this happens 50 times, it's 50*100 = 5000, 5000 + 2550 = 7550.

No need for formula, just simple number pattern here.

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Originally posted by rf3d3r3r on 28 Jan 2013, 11:19.
Last edited by rf3d3r3r on 16 Jun 2015, 22:01, edited 1 time in total.
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Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

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New post 03 Oct 2013, 03:44
I am totally confused. in the math book, the formula for the first n even number is n(n+1)

and here there is another formula used, mean*# of terms


Bunuel please explain!!!
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Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

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New post 03 Oct 2013, 05:33
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Aldossari wrote:
I am totally confused. in the math book, the formula for the first n even number is n(n+1)

and here there is another formula used, mean*# of terms


Bunuel please explain!!!


The sum of n first positive even numbers is n(n+1) --> the sum of first 50 positive even numbers is 50*51 = 2,550
The sum of evenly spaced set is (mean)*(# of terms) --> the sum of first 50 positive even numbers is (2+100)/2*50 = 51*50 = 2,550.
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Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

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New post 28 Oct 2014, 01:30
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2 + 4 + 6 + 8 ................... + 100 = 2550 ................. (1)

102 + 104 + 106 + 108 ............ + 200 = ?? ................... (2)

Just note that we have to add 100 to each term of series (1) to get the corresponding terms in series (2)

(2+100) + (4+100) + (6+100) + (8+100) ................ + (100+100)

This 100 ha to be added 50 times, so result of series (2) would be

2550 + 50*100 = 7550

Answer = B
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Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

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New post 01 Jul 2017, 15:20
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enigma123 wrote:
The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102 to 200 inclusive?

A. 5,100
B. 7,550
C. 10,100
D. 15,500
E. 20,100



The sum of the first 50 positive even integers is 2550.
In other words, 2 + 4 + 6 + 8 + ...+ 98 + 100 = 2550

We want the sum: 102 + 104 + 106 + . . . 198 + 200
IMPORTANT: Notice that each term in this sum is 100 GREATER than each term in the first sum.
In other words, 102 + 104 + 106 + . . . 198 + 200 is the SAME AS...
(100 + 2) + (100 + 4) + (100 + 6) + ... + (100 + 98) + (100 + 100)
We can rearrange these terms to get: (100 + 100 + ... + 100 + 100) + (2 + 4 + 6 + 8 + ...+ 98 + 100)

IMPORTANT: There are 50 100's in the red sum, and we're told that the blue sum = 2550

So, our sum = 50(100) + 2550
= 5000 + 2550
= 7550

Answer:

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Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

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New post 10 Jun 2018, 03:40
enigma123 wrote:
The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102 to 200 inclusive?

A. 5,100
B. 7,550
C. 10,100
D. 15,500
E. 20,100

For any EVENLY SPACED SET:
Count = \(\frac{(biggest - smallest)}{(increment)}\) + 1
Average = median = \(\frac{(biggest + smallest)}{2}\)
Sum = (count)(average)
The INCREMENT is the difference between successive values.


Even numbers between 102 and 200, inclusive:
Here, the integers are EVEN, so the increment = 2.
Count = \(\frac{(200-102)}{2}\) + 1 = 50.
Average = \(\frac{(200+102)}{2}\) = 151.
Sum = (50)(151) = 7750.




Alternate approach:

102, 104, 106......196, 198, 200.

The median is halfway between 102 and 200:
\(\frac{(102+200)}{2}\)= 151.
Of the 100 integers between 101 and 200, inclusive, half are odd, half are even.
Thus, in the set above, there are 25 even integers to the LEFT of 151 and 25 even integers to the RIGHT of 151, for a total of 50 even integers.

Add successive PAIRS of integers, working from the OUTSIDE IN:
102+200 = 302.
104+198 = 302.
106+196 = 302.

Notice that the sum of each pair = 302.
Since there will be 25 of these pairs, the following sum is yielded:
25*302 = 7550.


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Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

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New post 06 Nov 2018, 07:59
enigma123 wrote:
The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102 to 200 inclusive?

A. 5,100
B. 7,550
C. 10,100
D. 15,500
E. 20,100





SUM OF N TERMS = \(\frac{n}{2} (2a+(n-1)d)\)

\(a\) = 102 the first term
\(d\) = 2 distance
\(n\) = 50 how many terms to add up


\(\frac{50}{2} (2 (102)+(50-1)2)\) = 7550



another approach Sum = average x number of terms

Average = \(\frac{(200 + 102)}{2} = 151\)

Number of terms \(\frac{(200-102)}{2} + 1 = 50\)

Sum of Even integers 151 *50 = 7,550



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Re: The sum of the first 50 positive even integers is 2550. What is the  [#permalink]

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New post 16 Oct 2019, 14:53
[[(200-102)/2 ] +1]*(200+102)/2
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Re: The sum of the first 50 positive even integers is 2550. What is the   [#permalink] 16 Oct 2019, 14:53
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