Question

The 288th term of the series a,b,b,c,c,c,d,d,d,d,e,e,e,e,e,f,f,f,f,f,f... is

• A. u
• B. v
• C. w
• D. x

Hint:

Notice how many times each letter is repeated. Add up the repeat counts until the total first reaches 288.

Answer 1

The Correct Option is D

Step 1: Find the pattern:
The series is a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, f, ... The first letter a appears once, b appears twice, c appears three times, and so on.
So the \(n\)-th letter of the alphabet appears exactly \(n\) times in a row.

Step 2: Count how many terms are used up by each letter:
After the letter number \(n\) is finished, the number of terms used is \(1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}\).
We need the first \(n\) for which this total reaches at least 288.

Step 3: Try values of n:
For \(n = 23\): \(\frac{23 \times 24}{2} = 276\). So the first 276 terms use up the letters a to w.
For \(n = 24\): \(\frac{24 \times 25}{2} = 300\). So the letters a to x use 300 terms.

Step 4: Locate term number 288:
Terms 1 to 276 are taken by the letters a to w. The letter number 24 then fills terms 277 to 300.
Term 288 lies between 277 and 300, so it is the 24th letter.

Step 5: Name the letter:
Counting in the alphabet: u is 21st, v is 22nd, w is 23rd, x is 24th.
So the 288th term is x.

Step 6: Check the other options:
Option 1 (u) would need the 288th term to lie in the block for the 21st letter, but that block ends at \(\frac{21 \times 22}{2} = 231\). So u is wrong.
Option 2 (v) ends at \(\frac{22 \times 23}{2} = 253\), which is less than 288. Wrong.
Option 3 (w) ends at 276, which is still less than 288. Wrong.

Final Answer:
The 288th term is the 24th letter, which is x. This is option 4. \[ \boxed{\text{Option 4: } x} \]