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The sum of the first 100 positive integers is 5,050. What is the sum : Problem Solving (PS)
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Re: The sum of the first 100 positive integers is 5,050. What is the sum [#permalink]
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Sum of first n consecutive numbers, = n/2(n+1)
SUm of first 10 consecutive numbers = 5*11 = 55
SUm of first 100 consecutive numbers = 50*101 = 5050 (Given)
SUm of first 200 consecutive numbers = 100*201 = 20,100 (Option E)
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Re: The sum of the first 100 positive integers is 5,050. What is the sum [#permalink]
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Rashed12 wrote:
The sum of the first 100 positive integers is 5,050. What is the sum of the first 200 positive integers?

A. 10,000
B. 10,200
C. 15,050
D. 20,050
E. 20,100

PS85402.01


Another approach is to use the formula for the sum of the first positive n integers.
1 + 2 + 3 + 4 + .... + n = (n)(n+1)/2

So, the sum of the first 200 positive integers = (200)(200+1)/2
= (200)(201)/2
= (100)(201)
= 20,100

Answer: E

Cheers,
Brent
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Re: The sum of the first 100 positive integers is 5,050. What is the sum [#permalink]
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Hi All,

We're told that the sum of the first 100 positive integers is 5,050. We're asked for the sum of the first 200 positive integers. This question can be approached in a number of different ways - and since we're dealing with the sum of CONSECUTIVE integers, this question can be solved rather easily by BUNCHING.

To start, since we're dealing with the first 200 positive integers, we can form 100 'groups of 2'.... Let's start by adding the smallest and largest numbers....

1+200 = 201

Then the next smallest and next largest and so on...

2+199 = 201
3+198 = 201
Etc.

We'll end up with 100 'pairs' that equal 201, thus, the overall sum of all 200 numbers is (100)(201) = 20,100

Final Answer:

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Re: The sum of the first 100 positive integers is 5,050. What is the sum [#permalink]
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Rashed12 wrote:
The sum of the first 100 positive integers is 5,050. What is the sum of the first 200 positive integers?

A. 10,000
B. 10,200
C. 15,050
D. 20,050
E. 20,100

PS85402.01


We can use the formula: average x number = sum. To find the average, we add the first and last numbers and divide by 2. Thus, the sum of the first 200 integers is:

(1 + 200)/2 x 200 = 201 x 100 = 20,100

Alternate solution:

Since the sum of the first 100 positive integers (i.e., 1 to 100) is 5,050 and each of the next 100 positive integers (i.e., 101 to 200) is 100 more than its counterpart in the first 100 integers (101 is 100 more than 1, 102 is 100 more than 2, etc.), then the sum of the integers from 101 to 200 (inclusive) is 100 x 100 = 10,000 more than 5,050, i.e., 15,050. Therefore, the sum of the first 200 positive integers (i.e., 1 to 200) is 5,050 + 15,050 = 20,100.

Answer: E
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Re: The sum of the first 100 positive integers is 5,050. What is the sum [#permalink]
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MBA HOUSE KEY CONCEPT: Summation of an arithmetic progression

Formula: (a1 + an) n / 2

a1 = first term = 1

an = last term = 200

n number of terms = 200

(1 + 200) 200 / 2 = 20100

E :)
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The sum of the first 100 positive integers is 5,050. What is the sum [#permalink]
Rashed12 wrote:
The sum of the first 100 positive integers is 5,050. What is the sum of the first 200 positive integers?

A. 10,000
B. 10,200
C. 15,050
D. 20,050
E. 20,100

PS85402.01

METHOD - I

We can also use the \(mean (average)\) \(=\) \(\frac{Sum-of-all-Elements}{Number-of-Elements}\), wherein we are asked to find the sum of the elements

Here,
1. Number of elements \(=\) \(200\)
2. Mean, in this case is a equally spaced list \(=\) \(\frac{First + Last}{2}\) \(=\) \(\frac{1 + 200}{2}\) \(=\) \(\frac{201}{2}\)
3. Sum of the elements \(=\) \(\frac{201}{2}\) \(*\) \(200\) \(=\) \(20,100\)

METHOD - II

We can also directly apply the formula \(\frac{n*(n + 1)}{2}\) were \(n\) stands for number of elements. In this case \(n\) equals \(200\).

\(\frac{200 * (200 + 1)}{2} = \frac{200 * 201}{2} = 20,100\)

Ans. E

Originally posted by Pritishd on 18 Jul 2020, 07:19.
Last edited by Pritishd on 19 Jul 2020, 02:18, edited 3 times in total.
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Re: The sum of the first 100 positive integers is 5,050. What is the sum [#permalink]
Approach:

formula to calculate SUM of first N numbers: \(\frac{N(N+1)}{2}\)

This case: \(\frac{200*201}{2} \)= 20100

Option E
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Re: The sum of the first 100 positive integers is 5,050. What is the sum [#permalink]
Using the formula n(n + 1)/2 to find the sum of first n natural numbers is definitely the fastest and the easiest way to get the answer. However, one may use an alternate way.

If a student understands the concept 'In an AP, Mean = Median', one can reach the answer really fast. Since the question in context asks about sum of the first 200 positive integers, just take 1st 199 positive integers.
Since median of first 199 positive integers is 100, therefore, Sum = number of terms x Median = 199 x 100 = 19900.

Now, add the remaining 200 to it.
19900 + 200 = 20100


Hence, the answer is E. :)
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Re: The sum of the first 100 positive integers is 5,050. What is the sum [#permalink]
First Approach
Sum of first 100 positive integers = 5050. Now, 101 to 200, each term will be 100 more than a certain term in 1 to 100. For example, 101 is 100 more than 1, 102 is again 100 more than 2... And so on till 200 is 100 more than 100.... Thus, the sum of 101 to 200 will be the sum of 1 to 100 + 100*100 = 5050 +10000 = 15050
Sum of all 1 to 200 = 5050+15050 = 20,100.

Second Approach
It can be interpreted that integers are consecutive and thus question hints AP series. It will be better that students recollect all the necessary concepts and formulas related to the AP series.
Use the formula for the sum of the first positive n integers.
1 + 2 + 3 + 4 + ... + n = (n)(n+1)/2
Thus, 1+2+…+200 = 200(200+1)/2 = 20100
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Re: The sum of the first 100 positive integers is 5,050. What is the sum [#permalink]
Expert Reply
Summation of an arithmetic progression where the first term is 1 and the progression has 1 as the common difference the formula is:

N(N + 1)/2 ; where N is the number of terms

200(200 + 1)/2 = 100 x 201 = 20100

E
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Re: The sum of the first 100 positive integers is 5,050. What is the sum [#permalink]
BrentGMATPrepNow wrote:
Rashed12 wrote:
The sum of the first 100 positive integers is 5,050. What is the sum of the first 200 positive integers?

A. 10,000
B. 10,200
C. 15,050
D. 20,050
E. 20,100

PS85402.01


Another approach is to use the formula for the sum of the first positive n integers.
1 + 2 + 3 + 4 + .... + n = (n)(n+1)/2

So, the sum of the first 200 positive integers = (200)(200+1)/2
= (200)(201)/2
= (100)(201)
= 20,100

Answer: E

Cheers,
Brent


Hi BrentGMATPrepNow, since first 100 positive integers sum is given 5,050. So thinking only need to calculate sum of 101 ~ 200 and add to 5050. However it doesn's seem to arrive at the correct answer. Not sure which part has gone wrong? Thanks Brent

101 ~ 200 (200-101+1) has 98 terms and 101+200 = 301
98 * 301/2 = 14749 + 5050 = 19799
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Re: The sum of the first 100 positive integers is 5,050. What is the sum [#permalink]
Expert Reply
Kimberly77 wrote:
BrentGMATPrepNow wrote:
Rashed12 wrote:
The sum of the first 100 positive integers is 5,050. What is the sum of the first 200 positive integers?

A. 10,000
B. 10,200
C. 15,050
D. 20,050
E. 20,100

PS85402.01


Another approach is to use the formula for the sum of the first positive n integers.
1 + 2 + 3 + 4 + .... + n = (n)(n+1)/2

So, the sum of the first 200 positive integers = (200)(200+1)/2
= (200)(201)/2
= (100)(201)
= 20,100

Answer: E

Cheers,
Brent


Hi BrentGMATPrepNow, since first 100 positive integers sum is given 5,050. So thinking only need to calculate sum of 101 ~ 200 and add to 5050. However it doesn's seem to arrive at the correct answer. Not sure which part has gone wrong? Thanks Brent

101 ~ 200 (200-101+1) has 98 terms and 101+200 = 301
98 * 301/2 = 14749 + 5050 = 19799


Hi Kimberly77,

I'm going to give you a hint so that you can retry this question. There are MORE than 98 terms between 101 and 200, inclusive.

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Contact Rich at: Rich.C@empowergmat.com
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The sum of the first 100 positive integers is 5,050. What is the sum [#permalink]
Thanks @EMPOWERgmatRichC...my silly mistake of terms calculation when it should've been 100. Time for a break now and thanks again.
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