Let L and M be the sum and average of four consecutive odd numbers (a, b, c…

CONCEPT

Consecutive odd numbers and consecutive even numbers both form an arithmetic progression with common difference 2, so each set is fixed once its smallest term is fixed. For any such set, sum = (number of terms) × (average), and when the count of terms is odd the average equals the middle term.

Hence four consecutive odd numbers starting at a are a, a + 2, a + 4, a + 6, with sum 4a + 12 and average a + 3; three consecutive even numbers starting at x are x, x + 2, x + 4, with sum 3x + 6 and average x + 2.

APPLICATION

1. Let a be the smallest of the four consecutive odd numbers, so L = 4a + 12 and M = a + 3.

2. Let x be the smallest of the three consecutive even numbers, so P = 3x + 6 and Q = x + 2.

3. Translate the condition M = Q − 6 into a, x: a + 3 = (x + 2) − 6, which simplifies to a = x − 7.

4. Translate the condition P = L − 16 into a, x: 3x + 6 = (4a + 12) − 16, which simplifies to 3x + 10 = 4a.

5. Substitute a = x − 7 into 3x + 10 = 4a: 3x + 10 = 4(x − 7) = 4x − 28, so x = 38.

6. Back-substitute to get a = 38 − 7 = 31, and therefore M = a + 3 = 34.

CROSS-CHECK

A valid solution must respect the stated parities and satisfy both given relations, so rebuild the two sets from a = 31 and x = 38 and test them.

• a = 31 is odd and x = 38 is even, so the four consecutive odd numbers are 31, 33, 35, 37 and the three consecutive even numbers are 38, 40, 42.

• L = 31 + 33 + 35 + 37 = 136 and M = 136 ÷ 4 = 34; P = 38 + 40 + 42 = 120 and Q = 120 ÷ 3 = 40.

• M = Q − 6 becomes 34 = 40 − 6 and P = L − 16 becomes 120 = 136 − 16, so both relations hold and M = 34.