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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]

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21 Mar 2012, 12:29
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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

Always get these incorrect. How to appraoch?

I approached it this way

Total number of terms = 200-102+1= 99

As its an evenly spaced set average = 200+102/2 = 151.

Sum = 151 * 99. But that's not the answer.
[Reveal] Spoiler: OA

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Re: Sum of even integers [#permalink]

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21 Mar 2012, 12:34
Quote:
Always get these incorrect. How to appraoch?

I approached it this way

Total number of terms = 200-102+1= 99

As its an evenly spaced set average = 200+102/2 = 151.

Sum = 151 * 99. But that's not the answer.

Very close ... the answer is 151*50= 7550

Reason being only the even numbers count
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Re: The sum of the first 50 positive even integers is 2550. What [#permalink]

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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

Always get these incorrect. How to appraoch?

I approached it this way

Total number of terms = 200-102+1= 99

As its an evenly spaced set average = 200+102/2 = 151.

Sum = 151 * 99. But that's not the answer.

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

The sum = mean*# of terms = (200+102)/2*50 = 7,550.

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

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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 [#permalink]

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26 Jan 2013, 07:24
use formula

sum = n/2[2a + (n-1)d ]

a first term, n no of terms.
d difference..

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

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26 Jan 2013, 11:29
Sachin9 wrote:
use formula

sum = n/2[2a + (n-1)d ]

a first term, n no of terms.
d difference..

works like charm :D

AP Sum = n/2[1st term + last term]
which is equivalent to = no. of terms x average
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Re: The sum of the first 50 positive even integers is 2550. What [#permalink]

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28 Jan 2013, 09:08
From what i see the first part of the question has nothing to do with the problem solving itself..the average between the last term and the first term multiplied by the number of terms will give us the sum of all numbers..number of terms is equal to the average between the last even integer and first even integer plus 1 in this case;(200-102)/2+1=50..therefore the sum of the even integers is equal to 50*(102+200)*1/2=7550 hence B
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The sum of the first 50 positive even integers is 2550. What [#permalink]

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28 Jan 2013, 11:19
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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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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 [#permalink]

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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

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

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03 Oct 2013, 05:33
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

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 [#permalink]

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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

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

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13 May 2015, 07:56
yes , no need to use 2250

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

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Re: The sum of the first 50 positive even integers is 2550. What   [#permalink] 27 Aug 2016, 21:10
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