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The sum of the first 50 positive even integers is 2,550. What is the

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Bunuel
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The sum of the first 50 positive even integers is 2,550. What is the [#permalink] New post   30 May 2016, 13:37
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The sum of the first 50 positive even integers is 2,550. What is the sum of the first 50 odd integers?

(A) 2,000
(B) 2,497
(C) 2,500
(D) 2,501
(E) 2,549
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C
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FacelessMan
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Re: The sum of the first 50 positive even integers is 2,550. What is the [#permalink] New post   30 May 2016, 13:44
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2 ways to solve this problem :

Sol1 :
Sum of first n even numbers = n(n+1) = 2550
Sum of first n odd numbers = n^2 = 50*50 = 2500

Sol2 :
Now the sum of first 50 even numbers can be represented as 2 + 4 + 6 + .... 100 = 2550
Now the sum of first 50 odd numbers can be represented as 1 + 3 + 5 + ... 99 = (2-1) + (4-1) + (6-1) + .... (100 -1)
= (2+4+6+...100) -1 -1 -1 ... 50 times = 2550 - 50 = 2500

Hence C.
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BrentGMATPrepNow
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Re: The sum of the first 50 positive even integers is 2,550. What is the [#permalink] New post   16 Jun 2017, 06:56
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Bunuel wrote:
The sum of the first 50 positive even integers is 2,550. What is the sum of the first 50 odd integers?

(A) 2,000
(B) 2,497
(C) 2,500
(D) 2,501
(E) 2,549


A different approach:

Let's observe some sums of consecutive odd integers and look for a pattern

Sum of first 1 odd integers: 1 = 1
Sum of first 2 odd integers: 1 + 3 = 4
Sum of first 3 odd integers: 1 + 3 + 5 = 9
Sum of first 4 odd integers: 1 + 3 + 5 + 7 = 16
Sum of first 5 odd integers: 1 + 3 + 5 + 7 + 9 = 25
Sum of first 6 odd integers: 1 + 3 + 5 + 7 + 9 + 11 = 36
.
.
.
See the pattern?
.
.
.
Sum of first n odd integers: n²

So, the sum of the first 50 odd integers = 50² = 2500

Answer:
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C


Cheers,
Brent
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Re: The sum of the first 50 positive even integers is 2,550. What is the [#permalink] New post   30 May 2016, 14:06
I chose the solution similar to one presented by FacelessMan:

Completely ignore the information about the sum of the first 50 even integers since there is a quicker way to solve the problem:

Sum of the odd numbers of the first consecutive integers n is \(n^2\), thus sum of odd integers = \(50^2\) = 2500

Answer: C
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Re: The sum of the first 50 positive even integers is 2,550. What is the [#permalink] New post   16 Jun 2017, 09:06
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Bunuel wrote:
The sum of the first 50 positive even integers is 2,550. What is the sum of the first 50 odd integers?

(A) 2,000
(B) 2,497
(C) 2,500
(D) 2,501
(E) 2,549


The sum of the first n odd integers equals \(n^2\)

\(= 50^2\)

\(= 2500\)

Thus, answer will be (C)
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Re: The sum of the first 50 positive even integers is 2,550. What is the [#permalink] New post   16 Jun 2017, 12:42
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Sum of first 50 odd integers = sum of arithmetic series = average of the series * n

average = (first + last)/2 = (1+99)/2 =50
n = 50

==> Sum of 50 odd integers = 50*50 = 2500
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Re: The sum of the first 50 positive even integers is 2,550. What is the [#permalink] New post   18 Jun 2017, 09:34
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Bunuel wrote:
The sum of the first 50 positive even integers is 2,550. What is the sum of the first 50 odd integers?

(A) 2,000
(B) 2,497
(C) 2,500
(D) 2,501
(E) 2,549


Sum of first 50 positive even integers is \(\frac{(2 + 100)}{2}*(50)\), which =

(51)(50) = 2,550

Sum of first 50 positive odd integers is \(\frac{(1 + 99)}{2}*(50)\), which =

(50)(50). There's one extra 50 in the even integer sum.* 2,550 - 50 = 2500.

Answer (C).

*(Sum of first 50 positive integers) - (Sum of first 50 odd integers) =

(51)(50) - (50)(50) =

(50)(51-50) =

(50)(1) = 50 more in sum of even integers than in sum of odd integers. (2550 - 50) = 2500.
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Re: The sum of the first 50 positive even integers is 2,550. What is the [#permalink] New post   28 Apr 2018, 08:02
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Bunuel wrote:
The sum of the first 50 positive even integers is 2,550. What is the sum of the first 50 odd integers?

(A) 2,000
(B) 2,497
(C) 2,500
(D) 2,501
(E) 2,549


The sum of the first 50 positive EVEN integers is 2,550
So, 2 + 4 + 6 + . . . + 98 + 100 = 2550

What is the sum of the first 50 odd integers?
So, we want the value of 1 + 3 + 5 + . . . + 97 + 99

IMPORTANT: Notice that:
1 = 2 - 1
3 = 4 - 1
5 = 6 - 1
.
.
.
97 = 98 - 1
99 = 100 - 1

So, 1 + 3 + 5 + . . . + 97 + 99 = (2 - 1) + (4 - 1) + (6 - 1) + . . . + (98 - 1) + (100 - 1)
= [2 + 4 + 6 + . . . + 98 + 100 ] - 1 - 1 - 1 - 1 . . . . -1 - 1
= [2550 ] - 1 - 1 - 1 - 1 . . . . -1 - 1

ASIDE: How many 1s are we subtracting from 2550 ?
Well, there are 50 even integers from 2 to 100, so we are subtracting fifty 1's
So, we get:
1 + 3 + 5 + . . . + 97 + 99 = [2550 ] - 1 - 1 - 1 - 1 . . . . -1 - 1
= [2550 ] - 50
= 2500

Answer: C

Cheers,
Brent
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Re: The sum of the first 50 positive even integers is 2,550. What is the [#permalink] New post   21 Jul 2020, 08:18
Bunuel wrote:
The sum of the first 50 positive even integers is 2,550. What is the sum of the first 50 odd integers?

(A) 2,000
(B) 2,497
(C) 2,500
(D) 2,501
(E) 2,549


We are asked to find the sum of first 50 odd integers i.e. \(1 + 3 + 5 + 7 + .... + n\)

1. We will first find out \(n\)
2. We will use the derived form of \(mean = \frac{sum-of-elements}{number-of-elements}\) = mean * number of elements to find the sum

\(n\) \(=\) first term + (number of terms - 1) * common difference \(=\) \(1 + (50 - 1) * 2 = 99\)

\(mean = \frac{first-term + last-term}{2}\) (for an evenly spaced set of numbers) \(= \frac{1 + 99}{2} = 50\)

sum of elements \(= 50 * 50 = 2,500\)

Ans. C
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Re: The sum of the first 50 positive even integers is 2,550. What is the [#permalink] New post   08 Mar 2024, 11:42
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