the 288th term of the series a, b , b, c, c, c, d, d, d, d,....

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

Hint AP sum formula n(n+1)/2

chetan2u · Accepted Answer

hi,

a is one time, b,2 times so on...
so 288=n(n+1)/2..
0r n(n+1)= 576... now 24*25=600...23*24<576 so x= 24.. and 24th alphabet is X..ans D

MahmoudFawzy · Answer

there is 1 letter a, 2 letter b, 3 letter c ....etc. in the set.
and we have 26 letters.

total number of terms in the set is 1+2+3+4+......+24+25+26 =  \frac{26}{2} (1+26) = 13*27 = 351

to remove Z letter, 351 - 26 = 325 (still above 288)
to remove Y letter, 325 - 25 = 300 (still above 288)
to remove X letter, 300 - 24 = 276 (below 288) --> which means that the 288th term is within the X letter region.

so the answer is 'D'

EMPOWERgmatRichC · Answer

Hi ynaikavde,

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

Gonner · Answer

Rich, 
could you please eloborate whats the logic behind

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.

EMPOWERgmatRichC · Answer

Hi abhinavk48,

Since the prompt asks us what the 288th letter is, we have to find an easy way to figure out what the prior 287 letters are.

Looking at the pattern in the question, we know that....
"A" shows up 1 time
"B" shows up 2 times
"C" shows up 3 times
etc.

The answer choices tell us that the 288th letter is one of the following: U, V, W, X or Z. Thus, we KNOW that the letters A through T all show up in the sequence BEFORE the 288th letter. We need a quick way to figure out how many letters make up this big group....A way that is FASTER than just adding 1+2+3+4+.....+19+20.

Since "A" shows up 1 time and "T" shows up 20 times, 1+20 = 21 letters
Since "B" shows up 2 times and "S" shows up 19 times, 2+19 = 21 letters
This pattern continues through the rest of the letters (and happens 10 times in total)....

10(21) = the first 210 letters are A through T.

From here, we can add the remaining letters until we get up to the 288th letter.

GMAT assassins aren't born, they're made,
Rich

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the 288th term of the series a, b , b, c, c, c, d, d, d, d,....

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the 288th term of the series a, b , b, c, c, c, d, d, d, d,....

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