Question: For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

1. 10,100
2. 20,200
3. 22,650
4. 40,200
5. 45,150

“For any positive integer n, the sum of the first n positive integers”- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. GMAT quant section is concerned with the logical and rational abilities of the students in deciphering graphical data and solving quantitative problems. The students must select the appropriate answer by calculating the sum with proper mathematical estimation. The students must hold a better idea of mathematical calculations to solve GMAT Problem Solving questions. The GMAT Quant topic in the problem-solving part notes calculative mathematical problems that are required to be solved with qualitative skills. The candidates can polish up their skills by answering from the book “GMAT Official Guide 2021”.

Solution and Explanation:

Approach Solution 1:

The problem statement informs that:

Given:

• the sum of the first n positive integers equals n(n+1)/2

Find Out:

• the sum of all the even integers between 99 and 301

The set of even integers between 99 and 301 represents the set that is evenly spaced.
Therefore, even integers between 99 and 301 (i.e arithmetic progression) are 100, 102, 104, ...., and 300.
The aggregate of the elements in the evenly spaced set = the average of the set multiplied by the number of terms present in a whole.
Therefore, average of the set = (largest number + smallest number)/2
                                                 = (300+100)/2
                                                 =200

Therefore, the number of terms present in a whole= (largest number- smallest number)/2 +1
                                                                                   =(300-100)/2 + 1
                                                                                   = 101

Thus the sum of all the even integers between 99 and 301 = 200* 101
                                                                                                =20,200

Correct Answer: (B)

Approach Solution 2:

The problem statement informs that:

Given:

• the sum of the first n positive integers equals n(n+1)/2

Find Out:

• the sum of all the even integers between 99 and 301

By applying the formula of the sum of the first n positive integers: n(n+1)/2, we get,
100+102+...+300= 2(50+51+..+150)

Therefore, we can say that the sum of integers from 50 to 150 is equal to the sum of integers from 1 to 49 subtracted from the sum of integers from 1 to 150.
Therefore, 2(50+51+..+150)= 2* (150(150+1)/2-49(49+1)/2)= 20,200
Thus the sum of all the even integers between 99 and 301 = 20,200

Correct Answer: (B)

Approach Solution 3:

The problem statement informs that:

Given:

• the sum of the first n positive integers equals n(n+1)/2

Find Out:

• the sum of all the even integers between 99 and 301

This question deals to find the sum of even integers between 99 and 301 that is the question intends to get the sum of 100+102+104+....298+300
Therefore, as per the formula of the sum of the first n positive integers, we can say,
100+102+104+....298+300= 2(50+51+52+...+149+150)

Therefore, the number of integers between 50 to 150= 150 - 50 + 1 = 101
That is there are 101 integers from 50 to 150.

Therefore, to find out 2(50+51+52+...+149+150), let's sum up the values in pairs:

....50 + 51 + 52 +...+ 149 + 150
+150+ 149+ 148+...+ 51 + 50
—------------------------------------------
= ...200+ 200+ 200+...+ 200 + 200

There are nearly 101 integers of 200 altogether in this set.
Hence, the sum of all 101 integers = 101 x 200 = 20,200

Thus, the sum of all the even integers between 99 and 301 =20,200

Correct Answer: (B)

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