Recovery fix (even numbers): sum equals 190 Five consecutive even numbers sum to 190. What is the sum of the largest and the smallest numbers?

Introduction / Context:
The original stem mentioning “odd” with sum 190 is inconsistent (sum of five odd integers is odd). Applying the Recovery-First Policy, we minimally repair the item to “five consecutive even numbers,” which aligns with 190. Now we can solve using averages.

Given Data / Assumptions:

• Five consecutive even integers.
• Total sum S = 190.
• We need the sum of the smallest and largest.

Concept / Approach:
For any five consecutive even numbers, the middle number equals the average (S/5). The sequence can be written as m−4, m−2, m, m+2, m+4 (with m even). The extremes sum to (m−4) + (m+4) = 2m, which is also 2 × (average).

Step-by-Step Solution:
Compute average: m = S / 5 = 190 / 5 = 38.Numbers: 34, 36, 38, 40, 42.Smallest + largest = 34 + 42 = 76.Therefore, the required sum is 76.

Verification / Alternative check:
Check total: 34 + 36 + 38 + 40 + 42 = 190; extremes sum 76 as computed. Any five-term symmetric AP has extremes sum = 2 × average.

Why Other Options Are Wrong:
73, 75, 77 arise from using consecutive odds or incorrect averaging; they do not match the repaired even-number scenario consistent with S = 190.

Common Pitfalls:
Not repairing the stem; using 5 consecutive integers instead of even integers; miscomputing average as 190/4.

Final Answer:
76