There are seven consecutive positive even numbers. The average of the square…

Concept: Seven consecutive even numbers can be written as x, x+2, x+4, x+6, x+8, x+10, x+12, where x is the first (smallest) term and x+12 is the last (largest) term. When a condition relates the squares of two terms of such a sequence, expanding the squares converts the condition into a quadratic equation in x, which can be solved by factorisation (splitting the middle term).

Application:

1. Let the seven consecutive even numbers be x, x+2, x+4, x+6, x+8, x+10, x+12.

2. The average of the squares of the first (x) and the last (x+12) is 712, so (x2 + (x+12)2) / 2 = 712.

3. Multiplying both sides by 2: x2 + x2 + 24x + 144 = 1424.

4. Combining like terms: 2x2 + 24x - 1280 = 0, which simplifies to x2 + 12x - 640 = 0.

5. Factorising by splitting the middle term: x2 + 32x - 20x - 640 = 0, i.e., x(x + 32) - 20(x + 32) = 0, so (x + 32)(x - 20) = 0.

6. This gives x = 20 or x = -32. Since the numbers must be positive, x = 20.

7. The sequence is 20, 22, 24, 26, 28, 30, 32. The largest number is 32, so the second largest number is 30 (x + 10 = 20 + 10).

Cross-check: With the first number 20 and the last number 32, the average of their squares is (202 + 322) / 2 = (400 + 1024) / 2 = 712, which matches the given condition, confirming the sequence is correct.