IAS 2023 — CSAT Reasoning

2023mcqmedium

What is the middle term of the sequence Z, Z, Y, Y, Y, X, X, X, X, W, W, W, W, W, …., A?

Official previous-year question

Verified 2 Aug 2026.

Options

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Solution

Correct Option (B)

The given sequence is Z, Z, Y, Y, Y, X, X, X, X, W, W, W, W, W, …., A.

This pattern indicates that the number of times each letter appears is determined by its position in the reverse alphabetical order, starting with Z appearing twice.

Z (26th letter from A) appears 2 times.
Y (25th letter from A) appears 3 times.
X (24th letter from A) appears 4 times.
W (23rd letter from A) appears 5 times.

Following this progression, the number of occurrences for a letter at alphabetical position 'p' (where A=1, B=2, ..., Z=26) is given by the formula (26 - p) + 2.

Therefore, for the letter A (p=1), it appears (26 - 1) + 2 = 27 times.

The sequence of occurrences for the letters from Z to A is 2, 3, 4, ..., 27.

To find the total number of terms in the sequence, we sum these occurrences:

Total terms = 2 + 3 + 4 + ... + 27

This is the sum of an arithmetic progression from 1 to 27, minus the first term (1):

Sum(1 to 27) = $\frac{27(27 + 1)}{2} = \frac{27 \times 28}{2} = 27 \times 14 = 378$

Total terms = 378 - 1 = 377.

The middle term of a sequence with an odd number of terms (N) is at position $\frac{N + 1}{2}$ .

Middle term position = $\frac{377 + 1}{2} = \frac{378}{2} = 189$ th term.

Now, we identify the letter corresponding to the 189th term by cumulatively summing the occurrences from the beginning of the sequence (Z):

Z: 2 terms (Cumulative sum: 2)
Y: 3 terms (Cumulative sum: 2 + 3 = 5)
X: 4 terms (Cumulative sum: 5 + 4 = 9)
W: 5 terms (Cumulative sum: 9 + 5 = 14)
V: 6 terms (Cumulative sum: 14 + 6 = 20)
U: 7 terms (Cumulative sum: 20 + 7 = 27)
T: 8 terms (Cumulative sum: 27 + 8 = 35)
S: 9 terms (Cumulative sum: 35 + 9 = 44)
R: 10 terms (Cumulative sum: 44 + 10 = 54)
Q: 11 terms (Cumulative sum: 54 + 11 = 65)
P: 12 terms (Cumulative sum: 65 + 12 = 77)
O: 13 terms (Cumulative sum: 77 + 13 = 90)
N: 14 terms (Cumulative sum: 90 + 14 = 104)
M: 15 terms (Cumulative sum: 104 + 15 = 119)
L: 16 terms (Cumulative sum: 119 + 16 = 135)
K: 17 terms (Cumulative sum: 135 + 17 = 152)
J: 18 terms (Cumulative sum: 152 + 18 = 170)
I: 19 terms (Cumulative sum: 170 + 19 = 189)

The 189th term falls within the block of 'I's, as 'I' occupies positions 171 through 189 in the sequence. Therefore, the middle term is I.

Incorrect Options:

(A) H: The letter H appears after I in the reverse alphabetical order in the sequence. Based on the cumulative sum, the 'H' block would start after the 189th term (I). Thus, H is not the 189th term.
(C) J: The letter J appears before I in the reverse alphabetical order. The cumulative sum up to J is 170. This means J occupies terms up to the 170th position. The 189th term is beyond the 'J' block.
(D) M: The letter M appears significantly earlier in the sequence. The cumulative sum up to M ends at the 119th position. Therefore, M is not the 189th term.

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Solution

Correct Option (B)

The given sequence is Z, Z, Y, Y, Y, X, X, X, X, W, W, W, W, W, …., A.

This pattern indicates that the number of times each letter appears is determined by its position in the reverse alphabetical order, starting with Z appearing twice.

Z (26th letter from A) appears 2 times.
Y (25th letter from A) appears 3 times.
X (24th letter from A) appears 4 times.
W (23rd letter from A) appears 5 times.

Following this progression, the number of occurrences for a letter at alphabetical position 'p' (where A=1, B=2, ..., Z=26) is given by the formula (26 - p) + 2.

Therefore, for the letter A (p=1), it appears (26 - 1) + 2 = 27 times.

The sequence of occurrences for the letters from Z to A is 2, 3, 4, ..., 27.

To find the total number of terms in the sequence, we sum these occurrences:

Total terms = 2 + 3 + 4 + ... + 27

This is the sum of an arithmetic progression from 1 to 27, minus the first term (1):

Sum(1 to 27) = $\frac{27(27 + 1)}{2} = \frac{27 \times 28}{2} = 27 \times 14 = 378$

Total terms = 378 - 1 = 377.

The middle term of a sequence with an odd number of terms (N) is at position $\frac{N + 1}{2}$ .

Middle term position = $\frac{377 + 1}{2} = \frac{378}{2} = 189$ th term.

Now, we identify the letter corresponding to the 189th term by cumulatively summing the occurrences from the beginning of the sequence (Z):

Z: 2 terms (Cumulative sum: 2)
Y: 3 terms (Cumulative sum: 2 + 3 = 5)
X: 4 terms (Cumulative sum: 5 + 4 = 9)
W: 5 terms (Cumulative sum: 9 + 5 = 14)
V: 6 terms (Cumulative sum: 14 + 6 = 20)
U: 7 terms (Cumulative sum: 20 + 7 = 27)
T: 8 terms (Cumulative sum: 27 + 8 = 35)
S: 9 terms (Cumulative sum: 35 + 9 = 44)
R: 10 terms (Cumulative sum: 44 + 10 = 54)
Q: 11 terms (Cumulative sum: 54 + 11 = 65)
P: 12 terms (Cumulative sum: 65 + 12 = 77)
O: 13 terms (Cumulative sum: 77 + 13 = 90)
N: 14 terms (Cumulative sum: 90 + 14 = 104)
M: 15 terms (Cumulative sum: 104 + 15 = 119)
L: 16 terms (Cumulative sum: 119 + 16 = 135)
K: 17 terms (Cumulative sum: 135 + 17 = 152)
J: 18 terms (Cumulative sum: 152 + 18 = 170)
I: 19 terms (Cumulative sum: 170 + 19 = 189)

The 189th term falls within the block of 'I's, as 'I' occupies positions 171 through 189 in the sequence. Therefore, the middle term is I.

Incorrect Options:

(A) H: The letter H appears after I in the reverse alphabetical order in the sequence. Based on the cumulative sum, the 'H' block would start after the 189th term (I). Thus, H is not the 189th term.
(C) J: The letter J appears before I in the reverse alphabetical order. The cumulative sum up to J is 170. This means J occupies terms up to the 170th position. The 189th term is beyond the 'J' block.
(D) M: The letter M appears significantly earlier in the sequence. The cumulative sum up to M ends at the 119th position. Therefore, M is not the 189th term.

IAS 2023 — CSAT Reasoning

Options

Did you get this right?

Solution

Correct Option (B)

Incorrect Options:

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IAS 2023 — CSAT Reasoning

Options

Did you get this right?

Solution

Correct Option (B)

Incorrect Options:

Your note

IAS Reasoning in other years

More IAS Reasoning questions

Other CSAT topics for IAS

Comprehension

Maths

Keep practicing Reasoning