This post will give you some **GMAT practice problems** related to the post GMAT Combinatorics 1.0: Introduction. These will not be** GMAT type problems** as MyGuru takes a...

# GMAT Quant Strategies: Summing an Arithmetic Sequence

One of the more difficult **GMAT problem types** deals with summing an arithmetic sequence. Problem 157 in the 12th edition Official Guide is an example:

157. 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?

(A) 10,100

(B) 20,200

(C) 22,650

(D) 40,200

(E) 45,150

In this problem the GMAT provides us with an equation, and we can use that equation along with a bit of algebra to solve the problem. However, this might not always be the case, and no doubt in the future the test writers will introduce problems without helpful hints - indeed there are a couple in the 12th and 13th editions of the OG. Consequently, students should know how to do this type of problem in general.

First a definition:

**Arithmetic sequences** are ordered lists of numbers (on the GMAT these numbers are most likely integers) where the difference between adjacent numbers is constant.

The most common example on the GMAT are lists of consecutive integers. For example:

{1, 2, 3, 4, 5, 6}, or, in general, any list of the form {*x*, *x *+ 1, *x *+ 2, *x + *3...}

Arithmetic sequences are similar, but the difference between the terms can be anything. For example:

{3, 7, 10, 13, 16}, {1/2, 5/2, 9/2, 13/2}, {2, 4, 6, 8, 10, 12, 14}, or, in general {*x*, *x* + d, *x + *2d, *x* + 3d...}

Everything we need to know about an arithmetic sequence can be summed up in three pieces of information:

1. The first (or last) term

2. The difference between the terms

3. The number of terms

For the most part, the first two pieces of information are explicitly or implicitly stated. If we are summing even integers, the difference is two and the first and last terms must both be even regardless of the range stated in the problem. The third piece of information is something we generally have to derive for ourselves. I'm going to use a simple example to explain how to do this.

How many integers are there between 7 and 15, inclusive?

Intuitively, it seems like we should find the difference between the two numbers, 15 - 7 = 8, and that will be the number of terms.

If we take the brute force approach of actually listing the integers we find that the intuitive approach is off by 1: {7, 8, 9, 10, 11, 12, 13, 14, 15}. If you count the integers, you'll find that there are nine of them in the list. It might seem like we should just take the brute force approach every time we need to solve this sort of problem, but for larger sets this would be impractical.

So, why is the intuitive approach incorrect? It pretty simple - by subtracting the 7, we don't include it in the set, and the count is off by one. To compensate for this, we add one to the difference. If *x *is less than *y*, the number of integers between *x* and *y*, inclusive is

*y* - *x + *1

We have to modify this formula if we want to use it count an arithmetic sequence where the difference between terms is greater than 1. If d is the difference between the terms in the sequence, and *x* is less than *y*, then the number of terms in the sequence is

**( y - x)/d + 1**

We have to be a bit careful applying this formula. If we're asked to find the number of multiples of 3 between 2 and 34, our first term is not 2, it's 3, and our last term isn't 34, it's 33 - this is because we're only counting multiples of 3.

So, now that we know how to count the number of terms we can move on to finding the sum of the terms.

First we need a couple of definitions

1. The **average or arithmetic mean** of a set of numbers is

The sum of the terms/The number of terms

For example, the average of the first 5 positive even integers (2, 4, 6, 8, and 10) is

**(2 + 4 + 6 + 8 + 10)/5 = 30/5 = 6**

2. The **median** of a set of ordered numbers is the middle term of the set or the average of the two middle terms. That is, if an ordered set has 5 numbers, the 3rd term will be the median; if an ordered set has 6 numbers, the average of the 3rd and 4th terms will be the median. Two examples:

To find the median of the set {8, 17, 7, 13, 4}, we order the terms of the set from least to greatest (or vice versa) and find the middle term. The middle term of the set {4, 7, 8, 13, 17} is 8. So the median of the set is 8.

To find the median of the set {8, 17, 7, 13, 4, 31} we follow the same procedure, but now we take the average of the two middle terms. The average of the two middle terms of the set {4, 7, 8, 13, 17, 31} are 8 and 13. So the median of the set is (8 + 13)/2 = 10.5

Arithmetic sequences of numbers have the nifty property that the median and mean of the set are the same number.

**Mean of an arithmetic sequence = Median of an arithmetic sequence**

Furthermore the mean of an arithmetic sequence is just the average of the first and last term, and since the median is equal to the mean, the median is also the average of the first and last term. If this seems mysterious, remember that there is a geometric interpretation of the mean. The mean of two numbers on the number line is the midpoint, e.g. the mean of 1 and 7 is 4, because the distance from 1 to 4 is equal to the distance betwen 4 and 7. In an arithmetic sequence the midpoint between the first and last term is exactly the median.

Now let's derive the formula for the sum of an arithmetic sequence.

Recall that

**Mean = Sum of terms/Number of terms**

Therefore

**(Mean)(Number of terms) = Sum of terms**

So, if *x* is the first term of an arithmetic sequence, *y* is the last term, and d is the difference, the sum of the sequence is

**[( x + y)/2][((y - x/)d) + 1] = Sum of terms **

Now let's return to problem 157 from the 12th OG:

157. 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?

(A) 10,100

(B) 20,200

(C) 22,650

(D) 40,200

(E) 45,150

Because we are only concerned with the even integers, the first term of the sequence is 100 and the last term of the sequence is 300. The mean of the sequence is

(100 + 300)/2 = 400/2 = 200

The number of terms in the sequence is

((300 - 100)/2) + 1 = (200/2) + 1 = 100 + 1 = 101

Thus, the sum of the sequence is

(101)(200) = 20,200

The explanation is the OG is much more convoluted, with unnecessarily obscure notation. The method presented here is clean and quick. Try it for yourself.

1. Find the sum of the first 100 positive integers

2. Find the sum of all the multiples of 3 between 1 and 100

3. Find the sum of the multiples of 7 between 47 and 624

4. The sum of a set of 50 consecutive positive odd numbers is 172. Find the fist and last term of the sequence

I hope you enjoyed this post. I will post answers to the questions with explanations in a subsequent post. Please visit our GMAT Tutoring page to learn more about MyGuru's approach to test prep.

#### About the Author

John E. is a Senior MyGuru GMAT tutor in Chicago. He also tutors the GRE, SAT, ACT, and almost anything math related. He holds a B.S. in Mathematics and English from Indiana University, and a M.S. in Mathematics from UIC. He's scored in the 99th percentile on all of the above standardized tests, and has worked for MyGuru and other companies to develop proprietary test prep questions and explanations of difficult quantitative concepts to help students improve their GMAT scores.

He's currently working with MyGuru to develop proprietary test prep material focused on the most difficult GMAT question types.

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