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

• The sequence follows a pattern where the $n$-th letter (counting Z as 1, Y as 2, ..., A as 26) appears $n$ times.

• The total number of terms in the sequence is the sum of the first 26 natural numbers: $\frac{26 \times 27}{2} = 351$.

• The middle term of a sequence with 351 terms is the $\frac{351 + 1}{2} = 176$-th term.

• We need to find which letter contains the 176th term by calculating cumulative sums: the sum of terms up to 'K' (12th letter) is $\frac{12 \times 13}{2} = 78$.

• The sum of terms up to 'L' (11th letter) is $\frac{11 \times 12}{2} = 66$.

• The sum of terms up to 'M' (10th letter) is $\frac{10 \times 11}{2} = 55$.

• Let's check the cumulative count from the end (A=1, B=2...): Sum up to 'N' (13th from end) is $\frac{13 \times 14}{2} = 91$. Sum up to 'M' (12th from end) is $\frac{12 \times 13}{2} = 78$. Sum up to 'L' (11th from end) is $\frac{11 \times 12}{2} = 66$. Sum up to 'K' (10th from end) is $\frac{10 \times 11}{2} = 55$. Sum up to 'J' (9th from end) is $\frac{9 \times 10}{2} = 45$. Sum up to 'I' (8th from end) is $\frac{8 \times 9}{2} = 36$. Sum up to 'H' (7th from end) is $\frac{7 \times 8}{2} = 28$. Sum up to 'G' (6th from end) is $\frac{6 \times 7}{2} = 21$. Sum up to 'F' (5th from end) is $\frac{5 \times 6}{2} = 15$. Sum up to 'E' (4th from end) is $\frac{4 \times 5}{2} = 10$. Sum up to 'D' (3rd from end) is $\frac{3 \times 4}{2} = 6$. Sum up to 'C' (2nd from end) is $\frac{2 \times 3}{2} = 3$. Sum up to 'B' (1st from end) is $\frac{1 \times 2}{2} = 1$. This approach is complex. Let's stick to the forward cumulative sum.

• Cumulative sum up to letter $k$ (where Z=1, Y=2... A=26) is $S_k = \frac{k(k+1)}{2}$. We need the smallest $k$ such that $S_k \ge 176$.

• For $k=18$ (which corresponds to the 18th letter from Z, i.e., I): $S_{18} = \frac{18 \times 19}{2} = 171$. This is less than 176.

• For $k=19$ (which corresponds to the 19th letter from Z, i.e., H): $S_{19} = \frac{19 \times 20}{2} = 190$. This is greater than 176.

• Since the 176th term falls between the 172nd and 190th terms, it lies within the block of the 19th letter.

• The 19th letter from Z (Z=1, Y=2, ..., I=18, H=19) is H. Wait, let's re-verify the letter mapping.

• Z is 1, Y is 2, X is 3, W is 4, V is 5, U is 6, T is 7, S is 8, R is 9, Q is 10, P is 11, O is 12, N is 13, M is 14, L is 15, K is 16, J is 17, I is 18, H is 19.

• The 18th letter is I. The cumulative count up to I is 171.

• The 19th letter is H. The cumulative count up to H is 190.

• The 176th term is in the H block. Therefore, the middle term is H.