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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

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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

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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

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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

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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

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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

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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.
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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.

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