Various ways..
1) making use of the info in the question..
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, Sum of 101 to 200 will be sum of 1 to 100+100*100=5050+10000=15050..
Sum of all 1 to 200 = 5050+15050=20,100.
2) Average
Average of 1 to 200 is middle of 100 and 101, so 100.50, thus sum = 200*100.50=20,100.
3) Sum = \(200*\frac{1+200}{2}\)=200*100.50=20,100.

E
_________________

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)

GIVEN: 1 + 2 + 3 + 4 + 5 + .....+ 99 + 100 = 5050

We want to find the sum 1 + 2 + 3 + 4 + 5 + .....+ 199 + 200
Notice that 1 + 2 + 3 + 4 + 5 + .....+ 199 + 200 = 1 + 2 + 3 + 4 + 5 + .....+ 99 + 100 + (100 + 1) + (100 + 2) + (100 + 3) + (100 + 4) + . . . (100 + 99) + (100 + 100)
= (1 + 2 + 3 + 4 + 5 + .....+ 99 + 100) + (1 + 2 + 3 + 4 + 5 + .....+ 99 + 100) + (100 + 100 + 100 + 100 + ...100 + 100) [ note that there are one hundred 100's in the sum]
= (5050) + (5050) + (100)(100)
= 10,100 + 10,000
= 20,100

Answer: E

Cheers,
Brent
_________________

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 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:

GMAT assassins aren't born, they're made,
Rich
_________________

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
_________________

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

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
_________________

Approach:

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

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

Option E
_________________