2024 CSAT — Q46

The sequence is a concatenation of descending blocks. Reading the runs that end in $1$:

Track the cumulative term-count and block sums (a block $(k,\dots,1)$ has sum $\tfrac{k(k+1)}{2}$):

After the $(6,\dots)$ block we have $22$ terms, summing $1+1+3+6+10+15+21 = 57$. We need $6$ more terms, taken from the start of $(7,6,5,4,3,2,1)$: those are $7,6,5,4,3,2$, summing $27$.

Answer: (b) 84.

The trap is the block boundary: $28$ does not land at the end of a block. Decompose into descending runs, sum complete blocks up to the largest cumulative $\le 28$ (here $22$ terms, sum $57$), then add exactly the first few terms of the next block ($7,6,5,4,3,2$). Counting one term too many or too few is the entire difference between (a), (b), and (c). The opening $(1),(1)$ duplication must be honoured or the whole cumulative shifts — read the printed terms literally before generalising.