*What is the sum of the first 50 positive integers?** *

(A) 1000

(B) 1225

(C) 1275

(D) 1400

(E) 1450

This is a daunting question, perhaps akin to the question that asks about two trains traveling at varying rates headed in the same direction. Just like the speeding locomotive question, this type of question should not intimidate you – once you understand the concept.

**Think of it this Way**

If I add up the first number and last number, I get 1 + 50 = 51. If I begin working my way inwards, i.e. imagining a number line, I next add 49 + 2 = 51, then 48 + 3 = 51. You will notice that the sum is always 51. So, as long as I increase by one at the low range, and decrease by 1 from the upper range, the sum of the two numbers will always equal 51.

The next question I want to ask myself is, how many pairs of numbers are there in the first 50 integers? The logic is, if we pair numbers the way we did in the preceding paragraph, we always get 51. So, I’m asking myself, how many 51’s are there. Dividing 50 by 2, we get the number of pairs: 25. Therefore, we have to multiply 25 x 51 to get the sum of the sequence, which is 1275.

So, whenever we need to find a consecutive series, we simply add the first plus the last (e.g. 1 + 50), and then take the number of digits (e.g. 50) and divide by 2 (remember we are looking for the pairs). Next, I multiply this result (50/2) by the first and last (1 + 50), and I get 25 x 51 = 1275.

Let’s try that with another, easier problem: What is the sum of the numbers 1 – 10.

Adding first plus last (1 + 10), I get eleven. Then, I take the number of digits, 10, divide by 2, and get 5. Next, I simply multiply 11 x 5 = 55.

**Time for a Word Problem**

*Bob is training for a fitness competition. In order to increase his maximum number of pull-ups, he follows the following routine: he begins with 25 pull-ups, rests for thirty seconds, and then does 24 pull-ups and rests, dropping one pull-up each time (25, 24, 23, etc.) until his final set of 11 pull-ups. How many total pull-ups does Bob do? *

*(A) **55*

*(B) **150*

*(C) **270*

*(D) **275*

*(E) **325*

**Solution:**

You could furiously add up the numbers 25 through 11, or, you could find the sum of all the numbers 1 – 25, and then subtract from that the sum of 1 to 10 (remember, Bob stops at 11 pull-ups). To find the sum of 1 – 25, we add first + last = 26 x 25/2 = 325. Now, we subtract the sum of 1 – 10, which is 55 (see above). This gives us a total of 270 pull-ups (Bob is obviously pretty strong).

The best way to improve at sequences is practice. At the same time, there are other twists to sequences. In the coming week, I will deal with more advanced problems, and other helpful tricks to help you deal with a type of problem that should not be intimidating. If you’d like a review of the basics, read GRE Math Video Lessons: Series and Counting Basics to make sure you have a solid grasp of every level.

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Hello,

I figured out a formula for the sums of sequences based on averages. For any sequence n…m regularly spaced by some distance d: Subtract m from n and divide by d. If inclusive, add 1. If exclusive, subtract 1. Multiply this by the average of n and m. For example, 3+6+9+12+15+18+21=84=((21+3)/2)*((21-3)/3+1)

Asa Collier

Hi Asa,

That’s great that you were able to derive that formula on your own. It is 100% correct and is very helpful on questions where you have to sum a list of regularly spaced numbers. I’m sure you will do very well on the GRE quant 🙂

hey chris eagerly waiting here for ur advanced series blog………..would u like to tell me the chances of facing series problems on test day…… r they gonna be advanced or basics???? can i depend on ur magoosh lessons or present blogs (series) ????? if not then plz advice then plz advice other reliable sources……thnx in advance…..

Sanjoy,

The blog/product provides a pretty good foundation for these question types. But if you want more practice the Manhattan 5lbs. should have a few questions. In general, probably slightly less than 50% chance that you’ll see such a question test day. It could be relatively straightforward or very difficult, so practicing with tough questions is a good idea.

Best of luck 🙂

Hi,

I tried using the above approach for the sum of all integers from 45 to 155 inclusive

and i get the answer 10100

Kindly correct.

the calculation i did is as follows:

(155-45)/1 + 1 = 101 * ((155+45)/2) = 101*100=10100

Hi Chris,

I did your second word problem differently:

I added the first number plus the last number (so 25 +11) and multiplied by sum of all the numbers from 11 to 25 (ie, 15 numbers). Then I divided this by 2. Basically: 36(15)/2 and I arrived at the same answer of 270 pull-ups, without having to do the additional step of subtracting the sum of numbers 1 to 10.

Is it okay to do it this way? Is my reasoning correct?

Thanks!

Hi Ally,

Yes, that works perfectly fine as well. Both are efficient methods of getting the right answer. Good job on picking up on your method 🙂

Thanks Chris!

You’re welcome 🙂

Hi Chris,

In question 1, ‘first 50 integers’ could also mean 0 to 49 and then the answer could have been 1225?

Please clarify if I am missing something!

Thanks

This is a good point! We’ve just changed it to read “first 50 POSITIVE integers,” which was intended. Thank you for catching that, Tejas!

Hi Chris,

First of all, thank you for all the great work that you do. 2ndly, Would you be able to provide another example similar to the question where you subtracted set b from set a in the advanced series video!

Thanks in advance!

Shamila

Sure :)!

Let’s say Set A is made up off the even numbers between 100 and 200, inclusive. Set B is made up of the odd numbers between 100 and 200, inclusive. What is Set A – B?

Let me know when you get the answer and I’ll tell you if it is right :).

What is the answer?

I got an answer of 150, i.e., sum of even numbers from 100 – 200 is 7650 and sum of odd numbers from 100 – 200 is 7500. Therefore 7650-7500 = 150.

Hope I am right.

51

Hi Chris,

This question is a great workout for the brain! 🙂

I’m coming up with three possible answers when I look at the question in different ways.

So far I have calculated 50, 100 and 151 as answers! I wonder where am I getting lost! 😀

You are welcome! I’m glad the question provided a little brain workout :).

Without seeing your work it is hard for me to know exactly where you are getting lost. One thing you may want to do is add (25+12)/2 x 14 and then add the 11 at the very end, which gives you 270.

Hope that helps!

Hi Chris,

I would like to know the answer for Set A-B question . I got 150 as answer. Please explain how to do it ?

Hi Chris,

Could you please elaborate how to solve this question?

Thanks! 🙂

Hi Chris,

Can you go over the Set A and Set B problem? Also, when do you use the formula N(N+1)/2? Finally, can you explain how to find the sum of the odd integers from 1 to 199 and the sum of the even integers from 2 to 198?

Thanks Chris for sharing this post.

For no. 1 I did it using the formula [n(n+1)]/2 for sum of n numbers

(50*51)/2 = 1275

for no. 2 first I determined the number of sets using the arithmetic sequence formula

25 = 11 + (n-1) -1

n = 15

or (25-11) + 1 = 15

then sum of arithmetic sequence to get the no. of pullups:

15/2 [2*25 + (15-1) -1]

= 270

Hi Mohamed,

Yes, I used the formula too but in such a way that I dissected the logic behind it. I don’t like to give students the formulas without giving them the logic behind it. But, at the same time, I think I should have written it out clearly, the way you did, once I’d shown how it was derived. So [n(n+1)]/2 where n is the number of things you are adding up.