Question

The set of natural numbers is partitioned into subsets $S_1 = \{1\}$, $S_2 = \{2, 3\}$, $S_3 = \{4, 5, 6\}$, $S_4 = \{7, 8, 9, 10\}$ and so on. The sum of the elements of subset $S_{50}$ is:

• A. 61250
• B. 65525
• C. 42455
• D. 62525

Hint:

For partitioned sets with consecutive counts, use arithmetic series sum to locate first and last terms of the required subset.

Answer 1

The Correct Option is B

Pattern: $S_k$ contains $k$ consecutive natural numbers.
$S_1$ has 1 number, $S_2$ has 2 numbers, $S_3$ has 3 numbers, etc.
Total numbers before $S_{50}$ = $1 + 2 + 3 + + 49 = \frac{49 \times 50}{2} = 1225$.
So, the first element of $S_{50}$ is $(1225 + 1) = 1226$.
Since $S_{50}$ has 50 numbers, last element = $1226 + 49 = 1275$.
Sum of $S_{50}$ = $\frac{\text{first} + \text{last}}{2} \times \text{count} = \frac{1226 + 1275}{2} \times 50$.
= $\frac{2501}{2} \times 50 = 1250.5 \times 50 = 62525$. Wait — this gives 62525, but option says 65525 — recheck.
If indexing error: Total numbers before $S_{50}$ actually = sum from 1 to 49 = 1225 (correct). First = 1226, last = 1275, sum = $\frac{1226+1275}{2} \times 50 = 1250.5 \times 50 = 62525$. So the correct answer is 62525. Possibly misprint in options.