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What is the sum of all consecutive integers from 10 to 100, inclusive?

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What is the sum of all consecutive integers from 10 to 100, inclusive? [#permalink]

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New post 18 Oct 2016, 02:11
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A
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C
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Re: What is the sum of all consecutive integers from 10 to 100, inclusive? [#permalink]

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Bunuel wrote:
What is the sum of all consecutive integers from 10 to 100, inclusive?

A. 4550
B. 4950
C. 5005
D. 5500
E. 5555


This question can be solved in many ways.

But using this formula, you can calculate the answer in seconds


SUM= n/2(Starting number+ Ending number)

n: total numbers between the starting no. and ending no.

n= 100-10+1= 91

SUM= (91/2)*(10+100)
= 91*55
=5005

Hence, answer C
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Re: What is the sum of all consecutive integers from 10 to 100, inclusive? [#permalink]

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New post 18 Oct 2016, 02:32
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Sum of 1st 100 integers=100*101/2 = 5050

Sum of 1st 9 integers = 9*10/2 = 45

Answer = 5050-45 = 5005

C


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Re: What is the sum of all consecutive integers from 10 to 100, inclusive? [#permalink]

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New post 18 Oct 2016, 02:56
Bunuel wrote:
What is the sum of all consecutive integers from 10 to 100, inclusive?

A. 4550
B. 4950
C. 5005
D. 5500
E. 5555


Total numbers from 10 to 100 will be 100-10+1 = 91.

Since the numbers are consecutive then we'll get difference as 1 between any two numbers.. So this is Arithematic progression.

Sn = n/2(first number + last number) => 91/2(10+100) => 5005

IMO option C.

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What is the sum of all consecutive integers from 10 to 100, inclusive? [#permalink]

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New post 18 Oct 2016, 03:31
Bunuel wrote:
What is the sum of all consecutive integers from 10 to 100, inclusive?

A. 4550
B. 4950
C. 5005
D. 5500
E. 5555


Sum of all consecutive integers from 10 to 100 = Sum of all consecutive integers up to 100 - Sum of all consecutive integers up to 9


Sum of all consecutive integers from 10 to 100 = \(100*\frac{101}{2}\) - \(9*\frac{10}{2}\)

Sum of all consecutive integers from 10 to 100 = \(5050\) - \(45\)

Sum of all consecutive integers from 10 to 100 = \(5005\)

Hence correct answer will be (C)....
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Re: What is the sum of all consecutive integers from 10 to 100, inclusive? [#permalink]

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New post 19 Oct 2016, 01:56
Sum of first n consecutive integers = n*(n+1)/2

Sum of 100 consecutive integers = 100*101/2

Sum of first 9 integers (we have to find sum of 10 to 100 inclusive, so we have to consider till 9 only) = 9*10/2

100*101/2 - 9*10/2

= 5005

Option C.

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Re: What is the sum of all consecutive integers from 10 to 100, inclusive? [#permalink]

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New post 23 Oct 2016, 03:44
Bunuel wrote:
What is the sum of all consecutive integers from 10 to 100, inclusive?

A. 4550
B. 4950
C. 5005
D. 5500
E. 5555

It is an arithmetic progression with the first term being 10 and the last term being 100

Sum= n/2(2a + (n-1)d)

N being the no. of terms equals 91 (90 numbers from 11 to 100. 10 being additional one as it is inclusive)

Sum = 91/2(10*2 + (91-1)*1)
=5005

Answer = Option C

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Re: What is the sum of all consecutive integers from 10 to 100, inclusive? [#permalink]

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New post 24 Oct 2016, 18:36
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Expert's post
Bunuel wrote:
What is the sum of all consecutive integers from 10 to 100, inclusive?

A. 4550
B. 4950
C. 5005
D. 5500
E. 5555


We know that the general average formula is average = sum/quantity. This formula can be rearranged to isolate the sum of the values in the set: sum = average x quantity. This tells us that the sum of any set of numbers is equal to the number of terms in the set multiplied by the average of the numbers in the set.

Let’s start by determining how many terms are in the set.

quantity = 100 - 10 + 1 = 91

Next we can determine the average of the numbers from 10 to 100, inclusive. Since we are dealing with consecutive integers, which form an evenly spaced set of numbers, we can use the following formula to determine the average:

average = (1st number + last number)/2

average = (10 + 100)/2 = 110/2 = 55

Now, the sum can be calculated by means of the rearranged formula:

sum = 55 x 91 = 5,005

Answer: C
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Re: What is the sum of all consecutive integers from 10 to 100, inclusive? [#permalink]

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New post 26 Oct 2016, 19:16
Question can be solved using: sum/# = avg

#: 100-10+1 = 91
avg: we know that there are 91 numbers and the avg will occur at 91/2 = 45....rounded up to 46 --> 46th number in the lineup will be the avg. To find what number that is: x-10+1 = 46--> x = 55

Thus..the equation can be transformed into the following: SUM/91 = 55 --> SUM = 5005

Hence, C is correct.

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Re: What is the sum of all consecutive integers from 10 to 100, inclusive? [#permalink]

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New post 12 Oct 2017, 13:05
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This is how I've been getting them right. Idk if this is the most efficient way though.

N for consecutive numbers = (First Number - Last Number) +1

Sum of numbers = (First Number + Last Number) * N))/2

1. (100-10) +1 = 91

2. (100+10)*91))/2 = 5005 C

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Re: What is the sum of all consecutive integers from 10 to 100, inclusive?   [#permalink] 12 Oct 2017, 13:05
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What is the sum of all consecutive integers from 10 to 100, inclusive?

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