The variance of the numbers 8,21,34,47, …, 320 is _______.

Solution

Maths sir · 4 steps

This is an AP, so shift and scale it back to 1, 2, 3, …, 25 and use the ready-made variance.

1. Count the terms. The numbers form an AP with first term 8 and common difference 13. From 8 + 13(n-1) = 320 we get n - 1 = 24, so there are n = 25 terms.
2. Shift and scale. Each term is xₖ = 8 + 13(k-1) = 13k - 5 for k = 1, 2, …, 25. Subtracting a constant does not change variance. Multiplying by 13 multiplies variance by 13², so Var(x) = 169 × Var(1, 2, …, 25).

   Var(1, 2, …, n) = (n+1)(2n+1)/6 - ((n+1)/2)² = (n² - 1)/12

3. Put in numbers. With n = 25: (625 - 1)/12 = 624/12 = 52. Then Var(x) = 169 × 52 = 8788.
4. So. The variance is 8788.

The common slip: 676. Multiplying 52 by d = 13 instead of d² = 169 gives 676, not 8788. Scaling data by d scales variance by d².

The idea to keep. For any AP of n terms with common difference d, variance is d²(n²-1)/12. The first term does not matter at all.

Answer from the NTA final answer key. Solution by PrepWiser.