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# The sum of the first 100 even positive integers – divided by the sum o

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The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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23 Jan 2016, 07:09
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Question Stats:

61% (01:20) correct 39% (01:06) wrong based on 320 sessions

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The sum of the first 100 even positive integers – divided by the sum of the next 100 even positive integers – is equal to which of the following?

A. 1/3
B. 101/ 301
C. 51/151
D. 1/2
E. 1

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Re: The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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23 Jan 2016, 08:40
2
The first 100 positive numbers are
2, 4, 6, ....198, 200.
If we make 50 pairs made by the first and the last, second and the second last, we get:
2+200 + 4+198 + 6+196 + ... + 100+102
=50*(202)
=10100

same
202 - 400 = 30100

so 101/301

Ans B
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Posts: 6797
Re: The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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23 Jan 2016, 09:03
1
The sum of the first 100 even positive integers – divided by the sum of the next 100 even positive integers – is equal to which of the following?

A. 1/3
B. 101/ 301
C. 51/151
D. 1/2
E. 1

Hi,
i will tell you a simpler way of doing these Qs..

Remember sum of consecutive numbers can eb found by finding average of these numbers * total number..

sum of the first 100 even positive integers= first even integer=2, and 100th even integer=200..
$$sum = \frac{{2+200}}{2}*100=101*100$$..

sum of the next 100 even positive integers= first even integer=202, and 100th even integer=400..
$$sum = \frac{{202+400}}{2}*100=301*100$$..

answer $$=\frac{{101*100}}{{301*100}}= \frac{101}{301}$$

B
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Re: The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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09 Dec 2016, 07:01
3
1
Sum of the first even integers: n(n+1)

Sum of the first 100-even integers: 100(101) = 10100
Sum of the first 200-even integers: 200(201) = 40200
Sum of the the even integers between the first 100 and the first 200: 40200-10100 = 30100

--> 10100/30100
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The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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28 Jul 2017, 20:27
2
First 100 even integers: 2...200
Sum of first 100 even integers: (a1+an)n/2 = (2+200)*100/2=202*50
Next 100 even integers: 202~400
Sum of next 100 even integers: (a1+an)n/2 = (202+400)*100/2=602*50
Ratio of the sums: 202/402 = 101/301
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Re: The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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28 Jul 2017, 22:52
1
The sum of the first 100 even positive integers – divided by the sum of the next 100 even positive integers – is equal to which of the following?

A. 1/3
B. 101/ 301
C. 51/151
D. 1/2
E. 1

Sum of first 100 even positive integers = 2 + 4 +6 +8 +.... + 200 = 2(1+2+3+4+..... +100) = 2* (100*101/2) = 100*101
Sum of next 100 even positive integers = 202 + 204 +206 +..... +400 = 2 (101 + 102 + ... + 200) = 2 ((1+2+3+.....+200)-(1+2+3+...+100))
= 2 [200* 201/2 - 100*101/2] = 200*201 - 100*101

Ratio = 100*101/ (200*201 - 100*101)= 101/(402 - 101) = 101/301

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Re: The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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28 Jul 2017, 23:25
The sum of the first 100 even positive integers – divided by the sum of the next 100 even positive integers – is equal to which of the following?

A. 1/3
B. 101/ 301
C. 51/151
D. 1/2
E. 1

Ans(B) 101/301

Given that, p = (2+4+6.....+200 ) [First 100 Terms] / (202+204....+400) [Next 100 Terms]
p= 2(1+2+...+100) / 2 ( 101+ 102+200)
P= (1+2+....+100) / (101+102+...+200) ( Cancelling out 2)
After this we can, we can solve this using 2 methods.
Method1:
To calculate the sum of a continuous series, with a same difference between any two continuous numbers, simply take the average of first & last number and multiple it by the number of terms. For ex : 4+7+10+13= (average of 13&4, multiplied by 4) = 8.5*4 = 34
Now in the original problem,
p = Average of (1&100)*100 / Average of (101&200)*100
p=(1+100)/(101+200) (Cancelling out 2, not shown here)

Method2: [This is a formula dependent approach and lengthy one also]
For Equation, P= (1+2+....+100) / (101+102+...+200)
We can use the formula for the sum of first "n" natural numbers : S= n(n+1)/2. But this formula works when the series starts from 1.
So, using this formula, we can calculate the sum for the numerator part of the above equation.
But for denominator, first we need to calculate the some of 200 numbers and then subtract the sum of first 100.
And this double calculation makes this approach slightly lengthy.

Thanks!
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Re: The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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02 Nov 2017, 14:55
paidlukkha wrote:
The first 100 positive numbers are
2, 4, 6, ....198, 200.
If we make 50 pairs made by the first and the last, second and the second last, we get:
2+200 + 4+198 + 6+196 + ... + 100+102
=50*(202)
=10100

same
202 - 400 = 30100

so 101/301

Ans B

why don't we use 0?
isn't it even integer??
Math Expert
Joined: 02 Sep 2009
Posts: 49271
Re: The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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02 Nov 2017, 21:28
2
soodia wrote:
paidlukkha wrote:
The first 100 positive numbers are
2, 4, 6, ....198, 200.
If we make 50 pairs made by the first and the last, second and the second last, we get:
2+200 + 4+198 + 6+196 + ... + 100+102
=50*(202)
=10100

same
202 - 400 = 30100

so 101/301

Ans B

why don't we use 0?
isn't it even integer??

"The sum of the first 100 even positive integers..."

Yes, 0 is an even integer but it's neither positive nor negative.
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The sum of the first 100 even positive integers – divided by the sum o  [#permalink]

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05 May 2018, 11:14
Here I would try to solve this with the help of a formula.
'Sum of the first N even natural numbers = N(N+1)'

So, Sum of 1st 100 even numbers = 100(101), and Sum of the 1st 200 even numbers =200(201)
Therefore, the sum of the Next 100 even numbers 'Minus' the 1st 100 even numbers = 200(201)-100(101) which is equal to 30100 OR 100(301).

So, The sum of the first 100 even positive integers – divided by the sum of the next 100 even positive integers equal to

100(101)/ 100(301) =101/301.

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The sum of the first 100 even positive integers – divided by the sum o &nbs [#permalink] 05 May 2018, 11:14
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