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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## sequences ##### This topic has expert replies Newbie | Next Rank: 10 Posts Posts: 1 Joined: 28 Jun 2015 ### sequences by vvishal » Sun Jun 28, 2015 10:13 pm if the series is given as a,b,b,c,c,c,d,d,d,d .....then which term will be replicate as u? (a) 200 (b) 210 (c) 215 (d) 229 (e) 232 Master | Next Rank: 500 Posts Posts: 363 Joined: 17 Oct 2010 Thanked: 115 times Followed by:3 members by theCEO » Mon Jun 29, 2015 2:42 am vvishal wrote:if the series is given as a,b,b,c,c,c,d,d,d,d .....then which term will be replicate as u? (a) 200 (b) 210 (c) 215 (d) 229 (e) 232 question seems to be missing some information Master | Next Rank: 500 Posts Posts: 363 Joined: 17 Oct 2010 Thanked: 115 times Followed by:3 members by theCEO » Mon Jun 29, 2015 3:16 am vvishal wrote:if the series is given as a,b,b,c,c,c,d,d,d,d .....then which term will be replicate as u? (a) 200 (b) 210 (c) 215 (d) 229 (e) 232 A :sum=1 BB:sum=1+2=3 CCC:sum=1+2+3=6 DDDD:sum=1+2+3+4=10 the pattern is that we are summing consecutive integers the formuala is n*(n+1)/2 The letter U occupies the 21st position Therefore the last U will be the 21*22/2 = 231 term The first U will be the 231 - 21 + 1 = 211 term Therefore U will be the 211 to 231 term V will occupy the next 22 spots: 232 to 253 T will occupy 20 spots before the U starts: 191 to 210 ### GMAT/MBA Expert Elite Legendary Member Posts: 10346 Joined: 23 Jun 2013 Location: Palo Alto, CA Thanked: 2867 times Followed by:503 members GMAT Score:800 by Rich.C@EMPOWERgmat.com » Mon Jun 29, 2015 8:51 pm Hi All The posted questions looks rather similar to the following question: The 288th term of the series a, b , b, c, c, c, d, d, d, d, e, e, e, e, e, .... is A) u B) v C) w D) x E) z D GMAT assassins aren't born, they're made, Rich Contact Rich at Rich.C@empowergmat.com ### GMAT/MBA Expert Elite Legendary Member Posts: 10346 Joined: 23 Jun 2013 Location: Palo Alto, CA Thanked: 2867 times Followed by:503 members GMAT Score:800 by Rich.C@EMPOWERgmat.com » Mon Jun 29, 2015 9:07 pm Hi All, With this question, you can use the answer choices to your advantage as well as the Arithmetic concept "bunching." We're given a sequence of increasingly repetitive terms: A shows up 1 time B shows up 2 times C shows up 3 times Etc. From the answers, we know that the 288th letter is either U, V, W, X or Z. Working backwards in the alphabet.... Z is the 26th letter X is the 24th letter W is the 23rd letter V is the 22nd letter U is the 21st letter This means that the prior 20 letters of the alphabet will ALL show up in the pattern that is described in the prompt, so we can start by figuring out the TOTAL number of letters in that first sequence of 20 different letters. Here, we can use "bunching" A shows up 1 time B shows up 2 times C shows up 3 times ...... T shows up 20 times 1+20 = 21 letters 2+19 = 21 letters 3+18 = 21 letters Etc. This means that there will be 10 pairs of letters that fit this pattern. 10(21) = 210 letters. From here, we can just "add" groups of letters until we hit the 288th letter. U will show up 21 times 210 + 21 = 231 V will show up 22 times 231 + 22 = 253 W will show up 23 times 253 + 23 = 276 .....notice that we're "close" to the 288th letter.... X will show up 24 times.... 276 + 24 = 300....this means that the 277th through 300th letters are all X. The letter X is the 288th letter. Final Answer: D GMAT assassins aren't born, they're made, Rich Contact Rich at Rich.C@empowergmat.com Master | Next Rank: 500 Posts Posts: 154 Joined: 21 May 2014 Thanked: 8 times Followed by:1 members by talaangoshtari » Fri Jul 03, 2015 2:29 am I don't understand this part 1+20 = 21 letters 2+19 = 21 letters 3+18 = 21 letters MBA Admissions Consultant Posts: 2278 Joined: 11 Nov 2011 Location: New York Thanked: 660 times Followed by:266 members GMAT Score:770 by Jim@StratusPrep » Sat Jul 11, 2015 4:51 am He is describing how the sum of consecutive terms works. In a more simple structure, if you were to sum 1 + 2 + 3, you could sum 1 + 2 = 3 and then you have 3 x 2 = 6. Or, for 1 + 2 + 3 + 4 + 5, you have 1 + 4 = 5, 2 + 3 = 5, so 5(3) = 15. That can get a little confusing, so I just stick with the number of terms, which is 21 times the average of the first and last term. GMAT Answers provides a world class adaptive learning platform. -- Push button course navigation to simplify planning -- Daily assignments to fit your exam timeline -- Organized review that is tailored based on your abiility -- 1,000s of unique GMAT questions -- 100s of handwritten 'digital flip books' for OG questions -- 100% Free Trial and less than$20 per month after.
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by Rich.C@EMPOWERgmat.com » Sun Jul 12, 2015 11:31 am
Hi talaangoshtari,

When dealing with large-scale addition, you do NOT have to add the numbers in order.

For example, 1+2+3+4+5+6 is the same as 1+6 + 2+5 + 3+4. Notice in this second example that we have 3 "pairs" of 7, so (3)(7) = 21 (which is the same as 1+2+3+4+5+6).

In this question, we're dealing with a larger amount of 'data', but the concept is the same...

There is 1 "A"
There are 2 "Bs"
There are 3 "Cs",
Etc.

1 "A" and 20 "Ts" = 21 letters
2 "Bs" and 19 "Ss" = 21 letters
Etc.

GMAT assassins aren't born, they're made,
Rich
Contact Rich at Rich.C@empowergmat.com

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