The average of the squares of the first 48 natural numbers is

Correct — A. Use the sum-of-squares formula, then divide by how many numbers there are.

Σn² for the first n naturals = n(n+1)(2n+1)⁄6
Average = Σn²⁄n = (n+1)(2n+1)⁄6

Put n = 48:
(48+1)(96+1)⁄6 = (49 × 97)⁄6
49 × 97 = 4900 − 147 = 4753
4753⁄6 = 792.1666… = 792.17 → option (a)

The division is the deciding step: 6 × 792 = 4752, leaving a remainder of 1, so the decimal part is 1⁄6 = 0.1666…, which rounds up to .17.

Why the others are wrong

• (b)
  793.17 — 793.17 is 4759⁄6. It needs the product 49 × 97 to come out six too high — the slip of subtracting 141 from 4900 instead of 147.
• (c)
  794.17 — 794.17 is 4765⁄6, which needs 49 × 97 to be twelve too high. The formula is right in every one of these; only the multiplication moves.
• (d)
  791.17 — 791.17 is 4747⁄6, the product six too low. All four options sit one apart precisely so a mis-multiplied 49 × 97 still lands on an option.

Two standard results carry this topic.

Sum of the first n squares = n(n+1)(2n+1)⁄6. Divide by n and the average of those squares is (n+1)(2n+1)⁄6 — the n cancels, which is why the average is quicker than the sum.

For n = 48 that is 49 × 97 ⁄ 6 = 4753⁄6 = 792.17.

Do not confuse it with the average of the first n natural numbers, which is (n+1)⁄2 — here 24.5. Squaring first pushes the average far above the middle value, because large squares dominate.

Key facts

• The sum of the squares of the first n natural numbers is n(n+1)(2n+1)⁄6.
• The average of those squares is (n+1)(2n+1)⁄6.
• For n = 48 the average is 49 × 97 ⁄ 6 = 4753⁄6 = 792.1666…, or 792.17 to two decimals.
• 49 × 97 = 4753, reached fastest as 4900 − 147.

Common traps

• Reporting n(n+1)(2n+1)⁄6 as the average without dividing by n
• Rounding 792.1666… down to 792.16
• Reading the stem as the square of the average rather than the average of the squares

No directly related past PYQ was found.