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Re: The sum of the first 50 positive even integers is 2550. What [#permalink]
26 Jan 2013, 10:29

Sachin9 wrote:

use formula

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

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

works like charm :D

To add:

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]
28 Jan 2013, 08:08

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

Re: The sum of the first 50 positive even integers is 2550. What [#permalink]
28 Jan 2013, 10:19

2

This post received KUDOS

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 [#permalink]
03 Oct 2013, 04:33

Expert's post

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 [#permalink]
13 May 2015, 06:56

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]
13 May 2015, 06:56

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