There are five consecutive odd numbers. The difference between the square of the average of the first two odd numbers and the square of the average of the last two odd numbers is 396. What is the smallest odd number in this sequence?

Introduction / Context:
This question combines sequences, averages and quadratic expressions. You are given a set of five consecutive odd numbers and a condition involving the squares of averages of the first two and the last two numbers. From this information, you must determine the smallest odd number in the sequence. This tests algebraic modelling and careful handling of odd-number sequences in aptitude exams.

Given Data / Assumptions:

• There are five consecutive odd numbers.
• The difference between the square of the average of the first two odd numbers and the square of the average of the last two odd numbers is 396.
• We are asked to find the smallest odd number in the sequence.
• The numbers are consecutive odd integers, so they differ by 2 each time.

Concept / Approach:
If the first odd number is n, the five consecutive odd numbers can be written as n, n + 2, n + 4, n + 6 and n + 8. The average of the first two is (n + (n + 2)) / 2, and the average of the last two is ((n + 6) + (n + 8)) / 2. We then square these averages, form their difference and equate it to 396. This leads to a linear equation in n once we simplify using algebraic identities. Solving this gives the smallest odd number directly.

Step-by-Step Solution:
Step 1: Let the five consecutive odd numbers be n, n + 2, n + 4, n + 6 and n + 8. Step 2: The average of the first two numbers is (n + (n + 2)) / 2 = (2n + 2) / 2 = n + 1. Step 3: The average of the last two numbers is ((n + 6) + (n + 8)) / 2 = (2n + 14) / 2 = n + 7. Step 4: Their squares are (n + 1)^2 and (n + 7)^2. The difference of these squares is given as 396. Step 5: Since the later average is larger, we take (n + 7)^2 − (n + 1)^2 = 396. Step 6: Use the identity a^2 − b^2 = (a − b) * (a + b). Here a = n + 7 and b = n + 1, so a − b = 6 and a + b = (n + 7) + (n + 1) = 2n + 8. Step 7: Therefore, (n + 7)^2 − (n + 1)^2 = 6 * (2n + 8) = 12n + 48. Step 8: Set 12n + 48 = 396. Subtract 48 to get 12n = 348, then divide by 12 to get n = 29.

Verification / Alternative check:
If n = 29, the five consecutive odd numbers are 29, 31, 33, 35 and 37. The average of the first two is (29 + 31) / 2 = 60 / 2 = 30. The average of the last two is (35 + 37) / 2 = 72 / 2 = 36. Their squares are 30^2 = 900 and 36^2 = 1296. The difference is 1296 − 900 = 396, which matches the given condition. Therefore, n = 29 is correct and is indeed the smallest odd number in the sequence.

Why Other Options Are Wrong:
Option 27: If 27 were the smallest, the sequence would not satisfy the given relationship between the squared averages when checked.
Option 31 and 33: These values correspond to other members of the sequence but not to the smallest odd number satisfying the algebraic equation.
Option 25: This also fails when substituted, as the computed difference of squares will not be 396.

Common Pitfalls:
Some learners misinterpret the phrase “difference between” and may set up the equation with the smaller square minus the larger one, which would give a negative value and no meaningful solution for this problem. Others may forget that the numbers are odd and try using increments of 1. Writing the sequence explicitly as n, n + 2, n + 4, n + 6 and n + 8, and using the difference of squares identity, makes the solution straightforward.

Final Answer:
The smallest odd number in the sequence is 29.