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What is the sum of the odd integers from 45 to 65, inclusive?

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What is the sum of the odd integers from 45 to 65, inclusive?  [#permalink]

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New post 25 Jun 2015, 12:22
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What is the sum of the odd integers from 45 to 65, inclusive?

A. 495
B. 550
C. 555
D. 600
E. 605

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Re: What is the sum of the odd integers from 45 to 65, inclusive?  [#permalink]

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New post 27 Jun 2015, 12:20
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The mean is 55.

Sum=Mean(# of elements)

There are 11 odd numbers between 45-65 inclusive. 11*55=605

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Re: What is the sum of the odd integers from 45 to 65, inclusive?  [#permalink]

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New post 28 Jun 2015, 05:21
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reto wrote:
What is the sum of the odd integers from 45 to 65, inclusive?

A. 495
B. 550
C. 555
D. 600
E. 605


CONCEPT: Sum of first n positive odd Integers (all odd integers from 1 through 2n) = n^2

i.e. Sum of first 33 odd Integers from 1 through 65 (inclusive) =33^2

Also, Sum of first 22 odd Integers from 1 through 43 (inclusive) =22^2

i.e. Sum of all Odd Integers from 45 through 65 (inclusive) = 33^2 - 22^2 = (33+22)(33-22) = 55*11 = 605
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Re: What is the sum of the odd integers from 45 to 65, inclusive?  [#permalink]

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New post 12 Oct 2017, 14:25
1
Number of odd integers = N = (Last Odd # - First odd #)/2)) +1
Sum of odd numbers = (First Odd # + Last Odd #)*N))/2

1. (65-45)/2) + 1 = 11 = N

2. (45+65)*11))/2 = 605 E
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Re: What is the sum of the odd integers from 45 to 65, inclusive?  [#permalink]

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New post 16 Oct 2018, 01:35
Total number of odd integers between 45 and 65 inclusive= ((65-45)/2)+1=11
Now, we can use the following formula for Sum,
S=n /2 [2a+(n-1)d] [Here, n=11, a=45 and d=2]
So, sum = 11/2[2*45+(11-1)*2]=605 [E]
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Re: What is the sum of the odd integers from 45 to 65, inclusive? &nbs [#permalink] 16 Oct 2018, 01:35
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What is the sum of the odd integers from 45 to 65, inclusive?

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