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Which of the following is the sum of all the even numbers between 1 an

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Which of the following is the sum of all the even numbers between 1 an  [#permalink]

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New post 12 Sep 2015, 11:16
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Which of the following is the sum of all the even numbers between 1 and 99, inclusive?

A. 2550

B. 2450

C. 2600

D. 2499

E. 2652
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Which of the following is the sum of all the even numbers between 1 an  [#permalink]

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New post 12 Sep 2015, 14:27
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hiteshahire22 wrote:
Which of the following is the sum of all the even numbers between 1 and 99, inclusive?

A. 2550

B. 2450

C. 2600

D. 2499

E. 2652


Sum of an arithmetic series = Sn = \(\frac{n(2a+(n-1)d)}{2}\), where n = number of terms = number of even numbers between 1 and 99 = 49, d = difference between 2 consecutive even terms = 2, a = 1st term of the series = 2.

Thus Sn = \(\frac{49(2*2+(49-1)2)}{2}\) = 49*50 = a bit less than 50*50 = a bit less than 50*50 but with 0 as the unit's digit =B is the only option satisfying this.

B , 2450 is the correct answer.
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Re: Which of the following is the sum of all the even numbers between 1 an  [#permalink]

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New post 12 Sep 2015, 15:35
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Hi hiteshahire22,

This question can be solved with "bunching."

We're asked for the SUM of all of the EVEN integers from 1 to 99, so we really want the sum of 2, 4, 6, 8.....98.

If you take the smallest and biggest numbers, you have...
2+98 = 100

If you take the next smallest and next biggest numbers, you have...
4+96 = 100

This pattern will continue on, so we're going to have a bunch of 100s, BUT we have to be careful to make sure that we don't miss a number if it's "in the middle" and doesn't get paired up. Since we know that the sum of each pair is 100, we can 'jump ahead' to find the last few pairs...

44+56 = 100
46+54 = 100
48+52 = 100
There IS a middle number: 50; this number does NOT get paired up.

Since 48 is the 24th even integer, we know there are twenty-four 100s + one 50. 2400+50 = 2450

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Re: Which of the following is the sum of all the even numbers between 1 an  [#permalink]

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New post 12 Sep 2015, 17:07
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We need to find two things- Number of even terms and then Sum of all the terms.

Number of even terms= last even number - 1st even number/ spacing between two even numbers + 1. so, 98-2/2+1 = 49. now Sum of all the terms = Number of even terms * last number + 1st / 2 which is 49*99+1/2 = 2450
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Re: Which of the following is the sum of all the even numbers between 1 an  [#permalink]

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New post 09 Sep 2017, 00:18
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There are 2 Formulas we need to know

1 Sum of even numbers <=n where n is a natural number

in case N is odd then sum is given by {(n-1)(n+1)/4}

in case N is even then sum is given by n(n+2)/4

So in our case since n=99 is odd we have (98*100/4)= 2450
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Re: Which of the following is the sum of all the even numbers between 1 an  [#permalink]

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New post 09 Sep 2017, 00:27
49*2=98
---->49 Numbers
n/2(a1+an)

49(2+98)/2
4900/2
2450
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Re: Which of the following is the sum of all the even numbers between 1 an  [#permalink]

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Re: Which of the following is the sum of all the even numbers between 1 an   [#permalink] 28 Jan 2019, 21:36
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