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

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The sum of the first 50 positive even integers is 2550. What [#permalink] New post 06 Mar 2009, 13:02
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: the-sum-of-the-first-50-positive-even-integers-is-2550-what-129468.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Mar 2014, 13:53, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: PS: first 50 positive even integers [#permalink] New post 06 Mar 2009, 13:23
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The first 50 even integers will be 2,4,6,8....100. Sum of the first 50 even integers = n(+1) = 50 * 51 = 2550. (Infact this fix is not needed to be explicitly specified. Already implied in the statement sum of the first 50 even integers)

Sum of even integers between 102 to 200 inclusive will be

102 + 104 + 108 + ....200 (there will be 50 terms)

This can be written en

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

= 100 * 50 + (2 + 4 + 8 .....100)

= 5000 + 2550 = 7550.

Ans 7550.
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Re: PS: first 50 positive even integers [#permalink] New post 06 Mar 2009, 14:15
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Nice...I would have sat there for 5 minutes counting this.

+1.
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Re: PS: first 50 positive even integers [#permalink] New post 06 Mar 2009, 16:14
ALT way.
There are total 50 even numbers.
Total=50 * (102+200/2)=50 * 151=7550
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Re: PS: first 50 positive even integers [#permalink] New post 06 Mar 2009, 17:00
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Another way of solving it.

Sum of integers 102, 104, 106....200 is nothing but finding some of Arithematic Series.

Sum of AP series = n/2(first term + last term)

First term = 102
Last term = 200
number of terms n = 50.

So ans is 50/2 (102 + 200) = 25 * 302 = 7550. :)
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Re: PS: first 50 positive even integers [#permalink] New post 07 Mar 2009, 15:41
The formula is n/2 (2a + (n-1)d)

n is the no of digits, a is the first number and d is the difference in ourcase 2
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Re: PS: first 50 positive even integers [#permalink] New post 27 Mar 2009, 10:28
Thanks this is really helpful!
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Re: PS: first 50 positive even integers [#permalink] New post 14 Oct 2009, 07:58
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My approach
(100*101/2-50*51/2)*2=7550
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Short cut for this PS question? [#permalink] New post 14 Feb 2011, 14:26
The sum of the first 50 positive even integers is 2550. What is the sum of the even integers from 102-200 inclusive?

A) 5100
B) 7550
C) 10100
D) 15500
E) 20100
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Re: Short cut for this PS question? [#permalink] New post 14 Feb 2011, 15:39
I am actually brain dead.

151 * 50...

too much reviewing hard concepts is breaking my brain.
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Re: Short cut for this PS question? [#permalink] New post 14 Feb 2011, 23:09
I saw this question few days back in the forum;

1+2+3+4+5+6+....+49+50 = 2550

102+104+106+...+198+200 = 100+2+100+4+100+6+...+100+98+100+100

= 50*100 + 2+4+6+8+....+98+100
= 5000 + 2(1+2+3+4+5+....+49+50)
= 5000 + 2*2550
= 5000 + 5100
= 10100

Ans: "C"
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Re: Short cut for this PS question? [#permalink] New post 15 Feb 2011, 07:12
even numbers from 102 to 200=98/2+1=50
Sum= (200+102)x50/2=7550
So B is the ans.
Hope it helps.
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Re: Short cut for this PS question? [#permalink] New post 15 Feb 2011, 20:20
The question wordings are wrong. It has to be the sum of first 50 even positive integers is 2550.
In that case, the answer is B

Solution:

sum of first 100 integers = (100 + 101) / 2
= 5050

sum of first 50 integers = (50 + 51) / 2
= 1275

Hence, sum of integers from 51 to 100 = 5050 - 1275
= 3775

Now, sum of even integers from 102 to 200 inclusive = 102 + 104 + .... + 200
= 2(51 + 52 + 53 + ... + 100)
= 2 (3775)
= 7550
= B

Hope it helps.
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Re: PS: first 50 positive even integers [#permalink] New post 27 Feb 2011, 21:02
Although it defeats the purpose of the information presented in question, it is much simpler way to find out the sum.

Sum = (Ave) (number of terms)

Ave of an evenly spaced set = (First Term + Last Term) / 2
So, Ave = (102 + 200)/2
= 151

number of terms (inclusive) between an evenly spaced set = ((Last Term - First Term)/Difference between in each term) + 1
So, Number of terms N = ((200 - 98)/2)+1
= 50
(The diff between each of the two even integers is 2. In this particular case, you didn't even need to go thru this calculation since it is obvious that there are 50 even and 50 odd numbers in each 100. But the formula helps for other trickier questions)

Therefore, Sum = 151 * 50
= 7550

hope that helps!
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Re: PS: first 50 positive even integers [#permalink] New post 27 Feb 2011, 21:26
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We are already given the sum of first fifty even numbers and now need to find the sum of next 50. If you notice, each number in this case (sum of next 50) is 100 more than the old series (sum of first 50) and there are 50 such numbers, so the sum will be just the sum of first 50 numbers + 50*100 = 2550+5000 = 7550
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Re: PS: first 50 positive even integers [#permalink] New post 28 Feb 2011, 06:02
Sum of First 100(1,2,3,4..) natural number =5050


Sum of 102+104+106-----200 = 2(51+52+53+54+55......100)
=2*(sum of (1 to 100)-sum(1 to 50))
=2*(5050-1275)
=7550
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Re: PS: first 50 positive even integers [#permalink] New post 18 Jul 2011, 00:03
Should this even integer start with 0 ? is 0 an even integer ???? 0 + 2 + 4 .... 98? = 2550?

Also, please solve this problem in detail and if shortcut method if any ..
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Re: PS: first 50 positive even integers [#permalink] New post 19 Jul 2011, 09:37
The first piece of information is redundant here.
We need the sum of even integers from 102 to 200(inclusive).
1. To find the sum in such questions, use the formula:

AVG or AM = Sum/no. of terms

2. The set is an evenly spaced set. i.e., each consecutive number has a constant difference with the prev. (const. diff here is 2, since even integers)

3. For an evenly spaced set,
Properties:
1. AM or AVG = (first number + last number)/2 i.e., AM = Mid-term
2. AM = Median
Now, we can find Avg.
avg. = (102+200)/2 = 302/2 = 151
To find number of terms,
Use this formula:
No. of terms in a evenly spaced set

= [(Last term - First term)/constant difference] + 1

no. of terms = (200-102)/2 +1
= 50

Therefore,

Sum = Avg * no. of terms
= 151*50
= 7550
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Re: PS: first 50 positive even integers [#permalink] New post 20 Jul 2011, 03:11
nikhilsamuel89 wrote:
The first piece of information is redundant here.
We need the sum of even integers from 102 to 200(inclusive).
1. To find the sum in such questions, use the formula:

AVG or AM = Sum/no. of terms

2. The set is an evenly spaced set. i.e., each consecutive number has a constant difference with the prev. (const. diff here is 2, since even integers)

3. For an evenly spaced set,
Properties:
1. AM or AVG = (first number + last number)/2 i.e., AM = Mid-term
2. AM = Median
Now, we can find Avg.
avg. = (102+200)/2 = 302/2 = 151
To find number of terms,
Use this formula:
No. of terms in a evenly spaced set

= [(Last term - First term)/constant difference] + 1

no. of terms = (200-102)/2 +1
= 50

Therefore,

Sum = Avg * no. of terms
= 151*50
= 7550



Should this even integer start with 0 ? is 0 an even integer ???? 0 + 2 + 4 .... 98? = 2550?
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Re: PS: first 50 positive even integers [#permalink] New post 20 Jul 2011, 03:15
siddhans wrote:
nikhilsamuel89 wrote:
The first piece of information is redundant here.
We need the sum of even integers from 102 to 200(inclusive).
1. To find the sum in such questions, use the formula:

AVG or AM = Sum/no. of terms

2. The set is an evenly spaced set. i.e., each consecutive number has a constant difference with the prev. (const. diff here is 2, since even integers)

3. For an evenly spaced set,
Properties:
1. AM or AVG = (first number + last number)/2 i.e., AM = Mid-term
2. AM = Median
Now, we can find Avg.
avg. = (102+200)/2 = 302/2 = 151
To find number of terms,
Use this formula:
No. of terms in a evenly spaced set

= [(Last term - First term)/constant difference] + 1

no. of terms = (200-102)/2 +1
= 50

Therefore,

Sum = Avg * no. of terms
= 151*50
= 7550



Should this even integer start with 0 ? is 0 an even integer ???? 0 + 2 + 4 .... 98? = 2550?


Yes Zero is even integer
Re: PS: first 50 positive even integers   [#permalink] 20 Jul 2011, 03:15
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