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GMAT Practice Problems: Sequences

Here are eight practice problems on Sequences.

14, 23, 32, 41, 50, 59, …

1) In the sequence above, each term is 9 more than the previous term.  What is the 41st term of the sequence?

    (A) 360
    (B) 365
    (C) 369
    (D) 374
    (E) 383

2) What is the difference between the fourth and third terms of the sequence defined by a_n = 3^n - n^2

    (A) 18
    (B) 23
    (C) 47
    (D) 65
    (E) 83

3) Which of the following could be true of at least some of the terms of the sequence defined by

b_n = (2n - 1)(2n + 3)

I. divisible by 15

II. divisible by 18

III. divisible by 27

    (A) I only
    (B) II only
    (C) I and II only
    (D) I and III only
    (E) I, II, III

4) Let S be the set of all positive integers that, when divided by 8, have a remainder of 5.  What is the 76th number in this set?

    (A) 605
    (B) 608
    (C) 613
    (D) 616
    (E) 621

5) Let T be a sequence of the form a_n = a_1 + d*(n - 1).  If a_3 = 17 and a_19 = 65, find a_10.

    (A) 37
    (B) 38
    (C) 39
    (D) 40
    (E) 41

6) What is the sum of all the multiples of 20 from 160 to 840?

    (A) 14,000
    (B) 17,500
    (C) 18,060
    (D) 28,000
    (E) 35,000

7) A sequence is defined by s_n = (s_{n - 1} - 1)(s_{n - 2}) for n > 2, and it has the starting values of s_1 = 2 and s_2 = 3.  Find the value of s_6.

    (A) 25
    (B) 32
    (C) 36
    (D) 93
    (E) 279


9) In the set of positive integers from 1 to 500, what is the sum of all the odd multiples of 5?

    (A) 10,000
    (B) 12,500
    (C) 17,500
    (D) 22,500
    (E) 25,000

10) If a_n=(a_{n - 1})(a_{n - 3}) - (a_{n - 4}), a_1 = 1, a_2 = 2, a_3 = 3, and a_4 = 4, what is the value of a_6?

    (A) 7
    (B) 11
    (C) 19
    (D) 42
    (E) 130

Solutions will follow this article.



Sequences are a tricky topic on the GMAT Quant section.  In that linked article, I discuss sequence notation, which is a variant of function notation, and I discuss recursive sequences, that is, sequences in which each term is determined by the previous term or terms.


Arithmetic Sequences

An arithmetic sequence is a sequence in which we add some fixed amount to each term to get the next term.  Another way to say that is that, if we subtract any term from the following term, the difference will always be the same: this difference is called the common difference of the arithmetic sequence.   Let d be the common difference.  Then, in algebraic form, the terms of the arithmetic sequence would be


Recall that, say, for the 3rd term, the little subscript 3 is the index, that is, the position on the list.  For this particular sequence, every term equals the first term plus a factor times d, and that factor is always one less than the index.  This means we can write the general term as:


That is a very important formula, although, as always, don’t just memorize it; instead, remember the logic of the argument that leads up to it.

This is an important formula because any evenly spaced list is an arithmetic sequence.  The consecutive multiples of any factor form an arithmetic sequence.

The sum of the first n terms in an arithmetic sequence is given by the formula


That’s the sum of the first term and the last term, times half the number of items on the list.  You can also thing of that as the average of the first & last terms times the number of items on the list.  One special case is the sum of the first n integer, given by




If you had some insights while reading that first article on sequences or the section on arithmetic sequences, then take another look at the problems above before looking at the solutions below.  If you found this article helpful, or if you have an alternative solution for solving any of these problems, please let us know in the comments section below!



Practice problem explanations

1) This is an arithmetic sequence, with

a_1 = 14 and d = 9

Using the formula for the nth term, we find that:

a_40= 14 + 9*40 = 14 + 360 = 374.

Answer = (D).

2) This is general sequence, with an explicitly defined nth term.


Answer = (C).

3) First of all, if n = 8, then

b_8 = (16 – 1)(16 + 3) = 15*19

We don’t have to calculate that: clearly, whatever it is, it is divisible by 15.  Similarly, if n = 12, then

b_12  = (24 – 1)(24 + 3) = 23*27

Whatever that equals, it must be divisible by 27.  Thus, I & III are true.  Notice that, for any integer, 2n must be even, so both (2n – 1) and (2n + 3) are odd numbers, and their product must be odd.  Every term in this sequence is an odd number.  Now, no odd number can be divisible by an even number, because there is no factor of 2 in the odd number.  Therefore, no terms could possibly be divisible by 18.  Statement II is absolutely not true.

Answer = (D).

4) Think about the first few numbers on this list:

5, 13, 21, 29, 37, 45, …

Notice that 5 is the first number in S, because when 5 is divided by 8, the quotient is zero and the remainder is 5.

This is an arithmetic sequence with

a_1= 5 and d = 8

Using the general formula for the nth term of an arithmetic sequence, we have

a_76= 5 + 8*(75) = 5 + 4*(150) = 5 + 600 = 605

Answer = (A).

5) The formula in the first sentence tells us that this is an arithmetic sequence.  The first term and the common difference are unknown, but we can generate two equations from the values of the two terms given.


Subtract the first equation from the second, and we get 16d = 48, which means d = 3.  From the value of third term, we can see that first term must equal 11.   Therefore,


Answer = (B).

6) This is arithmetic sequence, and we have the first and last terms already.  How many terms are there?  Well, 160 = 20*(8) and 840 = 20*(42); we have to use inclusive counting to see that there are 42 – 8 + 1 = 35 terms.


Answer = (B).

7) This is a recursive sequence, and we have to find it term by term.


Answer = (E).

8) This is a recursive sequence, so we have to work backwards term by term.


Answer = (A).

9) Let’s think about the terms in this sequence:

5, 15, 25, 35, …., 485, 495

The first term is 5 and the last is 495.  There are 100 multiples of 5 from 1 to 500, so there are 50 odd multiples and 50 even multiple.  The sum is:


Answer = (B).

10) This recursive sequence is probably more difficult than anything that the GMAT is going to throw you, but solving this problem is not too bad.  As with any recursive sequence, we have to go term by term.


Answer = (C).


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8 Responses to GMAT Practice Problems: Sequences

  1. Meg August 24, 2016 at 12:00 pm #

    I got 365 for the first one. shouldn’t it be 39 times 9 since n-1 is within the parentheses, not just (n) for the formula?

    • Meg August 24, 2016 at 12:02 pm #

      I just answered my own question. It’s looking for 41st term, not 40th….oops.

  2. pushkar November 21, 2014 at 8:55 am #

    Hi Mike, i got 613 as an answer for question no 4…and when i read explanation i knew the answer should be 605. but still i am confused how 5 comes ahead and how when it is divided by 8 remainder is 5 …confused ..can u elaborate!! pls mike!!

    • Mike MᶜGarry
      Mike November 21, 2014 at 10:58 am #

      Dear Pushkar,
      I’m happy to respond. 🙂 Think about it this way. Of course, I think you will agree that, when 13 is divided by 8, the quotient is 1 and the remainder is 5. Thank about what this means. Suppose I walk into a room with eight children, and I have 13 cookies in a bag. Of course, I can give one cookie to each child, and 5 will remain in the bag (of course, I dare not mention these extra cookies to the children!) I walk out of the room with 5 in the bag: that’s my literal remainder.
      Now, think about 5 divided by 8. Suppose I walk into the same room full of eight children with only 5 cookies in my bag. Well, if I have any brains at all, I will not breathe a word of what’s in my bag to the children. Because I don’t have enough for every child, I am not going to give out any cookies. I walk out of the room with the same 5 cookies still in my bag — again, that’s the literal remainder.
      When we divide the 5 by 8, the 8 simply won’t go into it at all, which is why the integer quotient is zero, but that still leaves us with the untouched 5 as our remainder. You may find this blog helpful:
      Mike 🙂

      • brian January 13, 2015 at 10:50 pm #

        Problem 4 says: “Let S be the set of all positive integers”. Is 0 a positive integer?

      • Haritha July 7, 2016 at 8:37 pm #

        Hi Mike,

        As in the question they said when divided by 8 will leave a remainder 5. So simply i went by checking all the answer options which will be divisible by 8 and leaves a reminder 5. The only option which satisfies the remainder 5 is 605. Could you please confirm whether my approach is correct in answering the question ?

        • Magoosh Test Prep Expert
          Magoosh Test Prep Expert July 12, 2016 at 7:46 am #

          Hi Haritha,

          Happy to help! 🙂 While that worked out for you this time, that’s good, but what if all of the answers in the list were such that dividing by 8 left a remainder of 5? How would you determine which one was the answer in that case? That’s why it is much better to think of this in arithmetic sequence terms rather than simply brute force solving. I hope that makes sense! 🙂

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