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

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:

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

_________________________________ Please kudos me if you find my post useful

Re: The sum of the first 50 positive even integers is 2550. What is the
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03 Sep 2009, 06:58

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5

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

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.

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

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

Re: The sum of the first 50 positive even integers is 2550. What is the
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03 Oct 2013, 05:33

1

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.
_________________

Re: The sum of the first 50 positive even integers is 2550. What is the
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01 Jul 2017, 15:20

Top Contributor

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

Re: The sum of the first 50 positive even integers is 2550. What is the
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10 Jun 2018, 03:40

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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

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.

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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