The average of the squares of the first 47 natural numbers is

Correct — D. The average of the squares of the first n natural numbers is (n + 1)(2n + 1) ⁄ 6 — the sum n(n + 1)(2n + 1)⁄6 divided by n.

n = 47
n + 1 = 48, 2n + 1 = 95
48 × 95 = 4560
4560 ⁄ 6 = 760 → option (d)

The division is exact here, so there is no rounding to argue over.

Why the others are wrong

• (a)
  759 — 759 would need the product 4554 (759 × 6), but 48 × 95 is 4560. The answer is a whole number reached exactly, not a rounded-down 759.7.
• (b)
  761 — 761 × 6 = 4566, six more than 4560. The four options sit one apart on purpose, so the multiplication has to be carried out rather than estimated.
• (c)
  762 — 762 × 6 = 4572. A rough estimate of n²⁄3 gives about 736 for n = 47, which is nowhere near close enough to separate options spaced by a single unit.

Two standard sums cover almost everything SSC asks in this line:
1 + 2 + … + n = n(n + 1)⁄2
1² + 2² + … + n² = n(n + 1)(2n + 1)⁄6

An average is a sum divided by how many terms there are, and there are n of them. Dividing the second formula by n cancels the leading n and leaves (n + 1)(2n + 1)⁄6 — worth memorising in that form, because SSC asks for the average far more often than the sum.

For n = 47 that is 48 × 95 ⁄ 6, and 48 is already a multiple of 6, so the arithmetic is short.

Key facts

• 1² + 2² + … + n² = n(n + 1)(2n + 1)/6.
• The average of those squares is (n + 1)(2n + 1)/6.
• For n = 47 this is 48 × 95 / 6 = 760 exactly.
• The sum itself for n = 47 is 47 × 760 = 35,720.

Common traps

• Stopping at the sum, 35,720, instead of dividing by n to reach the average
• Writing 2n + 1 as 94 rather than 95
• Rounding when the division is exact and drifting to 759 or 761

No directly related past PYQ was found.