Five consecutive odd integers sum to 175 — Find the sum of the second-largest integer and the square of the smallest integer among them.

Introduction / Context:
Consecutive odd numbers form an arithmetic sequence with common difference 2. When the sum of five consecutive odd integers is given, the middle term equals the average. Once the sequence is identified, compute the requested combination: the second-largest integer plus the square of the smallest integer.

Given Data / Assumptions:

• Five consecutive odd integers have sum 175.
• Let the middle term be M; then the set is M - 4, M - 2, M, M + 2, M + 4.
• We need (second-largest) + (smallest)^2.

Concept / Approach:
For any five-term arithmetic progression, the sum equals 5 * (middle term). Hence M = 175 / 5 = 35. Then enumerate the integers and perform the required operations carefully to avoid arithmetic slips.

Step-by-Step Solution:
Find middle term: M = 175 / 5 = 35.List numbers: 31, 33, 35, 37, 39.Smallest = 31; second-largest = 37.Compute 31^2 = 961.Required sum = 37 + 961 = 998.

Verification / Alternative check:
The set indeed sums to 31 + 33 + 35 + 37 + 39 = 175. Squaring 31 gives 961, and adding 37 yields 998.

Why Other Options Are Wrong:

• 986 / 995 / 989 / 979: These values are close but result from mis-squaring 31, confusing “second-largest” with another term, or arithmetic mistakes in addition.

Common Pitfalls:
Taking the largest (39) instead of the second-largest (37); incorrectly computing 31^2; assuming the middle term is 34 or 36 due to parity mistakes.

Final Answer:
998