CSAT Solved Papers/ 2023/Q55

2023 CSAT — Q55

Quant Logical & quantitative reasoning 2.5 marks Hard

What is the middle term of the sequence

Z, Z, Y, Y, Y, X, X, X, X, W, W, W, W, W, \dots, A?

A H
B I Answer
C J
D M

Worked rationale

The blocks run with growing length: $Z$ appears $2$ times, $Y$ thrice, $X$ four times, $W$ five times — the $k$ -th letter (counting $Z$ as $k=1$ ) appears $k+1$ times, down to $A$ ( $k = 26$ , appearing $27$ times).

Total number of terms:

\sum_{k=1}^{26}(k+1) = \frac{26\cdot 27}{2} + 26 = 351 + 26 = 377.

The middle term of $377$ terms is the $\frac{377+1}{2} = 189$ th term.

Cumulative count through the first $j$ letters is $\sum_{k=1}^{j}(k+1) = \dfrac{j(j+3)}{2}$ . Find $j$ with the $189$ th term:

$j = 17$ : $\frac{17\cdot 20}{2} = 170$ .
$j = 18$ : $\frac{18\cdot 21}{2} = 189$ .

So terms $171$ – $189$ are the $18$ th letter, and the $189$ th term is its last. The $18$ th letter counting down from $Z$ is $Z, Y, X, W, V, U, T, S, R, Q, P, O, N, M, L, K, J, \mathbf{I}$ — I.

Answer: (b) I.

Visual solution

The same solve, worked by hand — read it, then trace it.

Hand-drawn worked solution for UPSC 2023 CSAT Q55 — Logical & quantitative reasoning — Tap the drawing to open it full size for the fine detail.

Why the other options miss

A

counted one too many: counts $19$ letters down (or miscounts the cumulative block boundary), landing one letter past $I$ .
C

counted one too few: stops at the $17$ th letter (cumulative $170 < 189$ ), one short of the block containing term $189$ .
D

an arithmetic slip: mis-totals the sequence length or the middle position, landing in a much earlier block.

Specialist insight

Two counts must be exact: the total ( $\sum (k+1) = 377$ , so the middle is the $189$ th) and the cumulative block boundary ( $\frac{j(j+3)}{2}$ , which hits $189$ exactly at $j = 18$ ). The closed forms beat tallying $377$ letters by hand. The lone trap is the alphabet index — letter $18$ down from $Z$ is $I$ , not the $18$ th letter up from $A$ ; confusing direction gives $R$ , and an off-by-one gives $H$ or $J$ .

The trap, in one line

$377$ terms $\Rightarrow$ middle is the $189$ th; cumulative $\frac{j(j+3)}{2} = 189$ at $j = 18$ , the $18$ th letter down from $Z$ is $I$ $\Rightarrow$ (b).

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