If the variance of the terms in an increasing A.P.,
b₁ , b₂ , b₃ ,....,b₁₁ is 90, then the common difference of this A.P. is_______.

Solution

Maths sir · 3 steps

Measure every term from the middle one, b6, and the variance comes out as a neat multiple of d squared.

1. Centre at b6. With 11 terms, the middle term b₆ is the mean of the A.P. So the deviations from the mean are -5d, -4d, …, 0, …, 4d, 5d.
2. Sum the squares. The squared deviations add up to 2d²(1 + 4 + 9 + 16 + 25) = 110d². Dividing by the 11 terms gives the variance.

   σ² = 110d²/11 = 10d² = 90

3. So. d² = 9, so d = ± 3. The A.P. is increasing, so d > 0 and the common difference is 3.

The common slip: 9. Stopping at d² = 9 gives 9. Take the square root, and use the word increasing to choose +3.

The idea to keep. For an A.P. of n terms with common difference d, the variance is (n²-1)d²/12. Adding a constant to every term never changes it, so the first term plays no part.

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