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For any positive integer n, the sum of the first n positive integers

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For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 09 Mar 2019, 14:20
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For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the integers between 250 and 350, inclusive?

(A) 100
(B) 6,600
(C) 20,200
(D) 30,300
(E) 55,000
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For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post Updated on: 09 Mar 2019, 15:51
Isn't the right answer 30.050?

Sum of the first 350 positive integers

\(\frac{350(350+1)}{2}=61.425\)

Sum of the first 250 positive integers

\(\frac{250(250+1)}{2}=31.375\)

Edit:

Doing the difference between the two:

\(62.750-31.375=30.050\)

Adding back the 250 who was inside the 31.375:

\(30.050+250=30.300\)

Originally posted by Mrpd on 09 Mar 2019, 14:36.
Last edited by Mrpd on 09 Mar 2019, 15:51, edited 1 time in total.
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 09 Mar 2019, 14:52
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Mrpd wrote:
Isn't the right answer 30.050?

Sum of the first 350 positive integers

\(\frac{350(350+1)}{2}=61.425\)

Sum of the first 250 positive integers

\(\frac{250(250+1)}{2}=31.375\)

Sum of all the integers between 250 and 350

\(62.750-31.375=30.050\)


How are you getting such small numbers?

Check this post by Bunuel: https://gmatclub.com/forum/for-any-posi ... l#p1046839
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 09 Mar 2019, 15:45
energetics wrote:
Mrpd wrote:
Isn't the right answer 30.050?

Sum of the first 350 positive integers

\(\frac{350(350+1)}{2}=61.425\)

Sum of the first 250 positive integers

\(\frac{250(250+1)}{2}=31.375\)

Sum of all the integers between 250 and 350

\(62.750-31.375=30.050\)


How are you getting such small numbers?

Check this post by Bunuel: https://gmatclub.com/forum/for-any-posi ... l#p1046839


Sorry, you're right.

I've excluded the 250 of the sum when I did \(62.750-31.375=30.050\).

The correct answer is 30.300 indeed.
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For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 08 Apr 2019, 15:44
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There is a quicker way than actually plugging in the integers here as the two posts above have done.

The formula \(\frac{n(n+1)}{2}\) can be used to take the sum of the integers from 1 to 350 inclusive, and then subtract the sum of the integers up to 249.

Rather than wasting time using the formula given, we can also find the sum of an integer set by using the formula \(mean * number of integers\). You can use this formula for any set of evenly spaced integers

From this, we know that the # of integers inclusive is 101 (251 to 350 is 100 integers, but we must also include 250 as part of the set). The mean can be easily found as 300.

Thus, we can multiply 101 * 300 and make quick mental math of it - (100 * 300 = 30,000) + (1 * 300) = 30,300 . Answer D
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Re: For any positive integer n, the sum of the first n positive integers  [#permalink]

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New post 13 Apr 2019, 13:45
Sum of n evenly-spaced integers of the form (a, a+d, a+2d, a+3d, ..., a+[n-1]d) is given by-

S = n/2 * [2*a + (n-1) *d]
where a is the first integer in the series,
d is the difference between two consecutive integers
n is the number of integers.

In the given question, total number of integers = (350-250+1)= 101
and d=1 (consecutive integers)
S= 101/2 * [2* 250 + 100 * 1]
S=30300
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Re: For any positive integer n, the sum of the first n positive integers   [#permalink] 13 Apr 2019, 13:45
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