- \n
- (A) 360<\/ol>\n<\/ol>\n
<\/p>\n
- \n
- (B) 365<\/ol>\n<\/ol>\n
<\/p>\n
- \n
- (C) 369<\/ol>\n<\/ol>\n
<\/p>\n
- \n
- (D) 374<\/ol>\n<\/ol>\n
<\/p>\n
- (E) 383<\/ol>\n
2) What is the difference between the fourth and third terms of the sequence defined by ![\"a_n](\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_986_f527bc3ae06e4a0d35816df6900f3952.png\" "\\"a_n")
<\/p>\n
- \n
- (A) 18<\/ol>\n<\/ol>\n
<\/p>\n
- \n
- (B) 23<\/ol>\n<\/ol>\n
<\/p>\n
- \n
- (C) 47<\/ol>\n<\/ol>\n
<\/p>\n
- \n
- (D) 65<\/ol>\n<\/ol>\n
<\/p>\n
- (E) 83<\/ol>\n
3) Which of the following could be true of at least some of the terms of the sequence defined by<\/p>\n
![\"b_n](\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_985.5_ab1eca8bc57fc612be96760828a157a5.png\" "\\"b_n")
<\/p>\n
I. divisible by 15<\/p>\n
II. divisible by 18<\/p>\n
III. divisible by 27<\/p>\n
- \n
- (A) I only<\/ol>\n<\/ol>\n
<\/p>\n
- \n
- (B) II only<\/ol>\n<\/ol>\n
<\/p>\n
- \n
- (C) I and II only<\/ol>\n<\/ol>\n
<\/p>\n
- \n
- (D) I and III only<\/ol>\n<\/ol>\n
<\/p>\n
- (E) I, II, III<\/ol>\n
4) Let S be the set of all positive integers that, when divided by 8, have a remainder of 5.\u00a0 What is the 76th<\/sup> number in this set?<\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n 5) Let T be a sequence of the form <\/p>\n <\/p>\n <\/p>\n <\/p>\n 6) What is the sum of all the multiples of 20 from 160 to 840?<\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n 7) A sequence is defined by <\/p>\n <\/p>\n <\/p>\n <\/p>\n 9) In the set of positive integers from 1 to 500, what is the sum of all the odd multiples of 5?<\/p>\n <\/p>\n <\/p>\n <\/p>\n <\/p>\n 10) If <\/p>\n <\/p>\n <\/p>\n <\/p>\n Solutions will follow this article.<\/p>\n <\/p>\n Sequences<\/a> are a tricky topic on the GMAT Quant section.\u00a0 In that linked article, I discuss sequence notation, which is a variant of function notation<\/a>, and I discuss recursive sequences, that is, sequences in which each term is determined by the previous term or terms.<\/p>\n <\/p>\n An arithmetic sequence<\/strong> is a sequence in which we add some fixed amount to each term to get the next term.\u00a0 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<\/strong> of the arithmetic sequence.\u00a0\u00a0 Let d be the common difference.\u00a0 Then, in algebraic form, the terms of the arithmetic sequence would be<\/p>\n Recall that, say, for the 3rd<\/sup> term, the little subscript 3 is the index<\/strong>, that is, the position on the list.\u00a0 For this particular sequence, every term equals the first term plus a factor times d<\/strong>, and that factor is always one less than the index.\u00a0 This means we can write the general term as:<\/p>\n That is a very important formula, although, as always, don\u2019t just memorize it; instead, remember the logic of the argument that leads up to it.<\/p>\n This is an important formula because any evenly spaced list is an arithmetic sequence.\u00a0 The consecutive multiples of any factor form an arithmetic sequence.<\/p>\n The sum of the first n terms in an arithmetic sequence is given by the formula<\/p>\n That’s the sum of the first term and the last term, times half the number of items on the list.\u00a0 You can also thing of that as the average of the first & last terms times the number of items on the list.\u00a0 One special case is the sum of the first n integer, given by<\/p>\n <\/p>\n 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.\u00a0 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!<\/p>\n <\/p>\n 1) This is an arithmetic sequence, with<\/p>\n Using the formula for the nth term, we find that:<\/p>\n Answer = (B)<\/strong>.<\/p>\n 2) This is general sequence, with an explicitly defined nth term.<\/p>\n Answer = (C)<\/strong>.<\/p>\n 3) First of all, if n = 8, then<\/p>\n We don\u2019t have to calculate that: clearly, whatever it is, it is divisible by 15.\u00a0 Similarly, if n = 12, then<\/p>\n Whatever that equals, it must be divisible by 27.\u00a0 Thus, I & III are true.\u00a0 Notice that, for any integer, 2n must be even, so both (2n \u2013 1) and (2n + 3) are odd numbers, and their product must be odd.\u00a0 Every term in this sequence is an odd number.\u00a0 Now, no odd number can be divisible by an even number, because there is no factor of 2 in the odd number.\u00a0 Therefore, no terms could possibly be divisible by 18.\u00a0 Statement II is absolutely not true.<\/p>\n Answer = (D)<\/strong>.<\/p>\n 4) Think about the first few numbers on this list:<\/p>\n 5, 13, 21, 29, 37, 45, …<\/p>\n Notice that 5 is the first number in S, because when 5 is divided by 8, the quotient is zero and the remainder<\/a> is 5.<\/p>\n This is an arithmetic sequence with<\/p>\n Using the general formula for the nth term of an arithmetic sequence, we have<\/p>\n Answer = (A)<\/strong>.<\/p>\n 5) The formula in the first sentence tells us that this is an arithmetic sequence.\u00a0 The first term and the common difference are unknown, but we can generate two equations from the values of the two terms given.<\/p>\n Subtract the first equation from the second, and we get 16d = 48, which means d = 3.\u00a0 From the value of third term, we can see that first term must equal 11.\u00a0\u00a0 Therefore,<\/p>\n Answer = (B)<\/strong>.<\/p>\n 6) This is arithmetic sequence, and we have the first and last terms already.\u00a0 How many terms are there?\u00a0 Well, 160 = 20*(8) and 840 = 20*(42); we have to use inclusive counting<\/a> to see that there are 42 \u2013 8 + 1 = 35 terms.<\/p>\n Answer = (B)<\/strong>.<\/p>\n 7) This is a recursive sequence, and we have to find it term by term.<\/p>\n Answer = (E)<\/strong>.<\/p>\n 8) This is a recursive sequence, so we have to work backwards term by term.<\/p>\n Answer = (A)<\/strong>.<\/p>\n 9) Let\u2019s think about the terms in this sequence:<\/p>\n 5, 15, 25, 35, …., 485, 495<\/p>\n The first term is 5 and the last is 495.\u00a0 There are 100 multiples of 5 from 1 to 500, so there are 50 odd multiples and 50 even multiple.\u00a0 The sum is:<\/p>\n\n
(A) 605<\/ol>\n<\/ol>\n
\n
(B) 608<\/ol>\n<\/ol>\n
\n
(C) 613<\/ol>\n<\/ol>\n
\n
(D) 616<\/ol>\n<\/ol>\n
(E) 621<\/ol>\n
![\"a_n](\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_985.5_b57f070ab13a6720356e388cbc3b2107.png\" "\\"a_n")
.\u00a0 If ![\"a_3](\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_986_211377b746aac9babe157808f3edb131.png\" "\\"a_3")
and ![\"a_19](\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_986_3b1e035af543fe4b3a06f5396663f3a9.png\" "\\"a_19")
, find ![\"a_10\"](\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_985.5_07f0a79aa5628bdc7716a977a48706e4.png\" "\\"a_10\\"\/")
.<\/p>\n\n
(A) 37<\/ol>\n<\/ol>\n
\n
(B) 38<\/ol>\n<\/ol>\n
\n
(C) 39<\/ol>\n<\/ol>\n
\n
(D) 40<\/ol>\n<\/ol>\n
(E) 41<\/ol>\n
\n
(A) 14,000<\/ol>\n<\/ol>\n
\n
(B) 17,500<\/ol>\n<\/ol>\n
\n
(C) 18,060<\/ol>\n<\/ol>\n
\n
(D) 28,000<\/ol>\n<\/ol>\n
(E) 35,000<\/ol>\n
![\"s_n](\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_983.5_90343da574a9f9ae8c95ea88f722d257.png\" "\\"s_n")
for n > 2, and it has the starting values of ![\"s_1](\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_986_a157387f34093bf3a02b5709ee023cf7.png\" "\\"s_1")
and ![\"s_2](\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_986_b9a9b0afeece5747f56041f0cd49d38b.png\" "\\"s_2")
.\u00a0 Find the value of ![\"s_6\"](\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_985.5_16b71dce4fcbfa76d904272b8a655ecc.png\" "\\"s_6\\"\/")
.<\/p>\n\n
(A) 25<\/ol>\n<\/ol>\n
\n
(B) 32<\/ol>\n<\/ol>\n
\n
(C) 36<\/ol>\n<\/ol>\n
\n
(D) 93<\/ol>\n<\/ol>\n
(E) 279<\/ol>\n
![\"gpp-s_img1\"](\"https:\/\/magoosh.com\/gmat\/files\/2014\/09\/gpp-s_img1.png\")
<\/a><\/p>\n\n
(A) 10,000<\/ol>\n<\/ol>\n
\n
(B) 12,500<\/ol>\n<\/ol>\n
\n
(C) 17,500<\/ol>\n<\/ol>\n
\n
(D) 22,500<\/ol>\n<\/ol>\n
(E) 25,000<\/ol>\n
=, , , , and , what is the value of ?<\/p>\n\n
(A) 7<\/ol>\n<\/ol>\n
\n
(B) 11<\/ol>\n<\/ol>\n
\n
(C) 19<\/ol>\n<\/ol>\n
\n
(D) 42<\/ol>\n<\/ol>\n
(E) 130<\/ol>\n
Sequences<\/h2>\n
Arithmetic Sequences<\/h2>\n
<\/a><\/p>\n<\/a><\/p>\n<\/a><\/p>\n<\/a><\/p>\nSummary<\/h2>\n
<\/a><\/p>\nPractice problem explanations<\/h2>\n
\u00a0= 14 and d = 9<\/p>\n= 14 + 9*39 = 14 + 351 = 365.<\/p>\n<\/a><\/p>\n = (16 \u2013 1)(16 + 3) = 15*19<\/p>\n \u00a0= (24 \u2013 1)(24 + 3) = 23*27<\/p>\n= 5 and d = 8<\/p>\n= 5 + 8*(75) = 5 + 4*(150) = 5 + 600 = 605<\/p>\n<\/a><\/p>\n<\/a><\/p>\n<\/a><\/p>\n<\/a><\/p>\n<\/a><\/p>\n