Let the six numbers a₁,a₂,a₃,a₄,a₅,a₆, be in A.P. and a₁+a₃=10. If the mean of these six numbers is 19/2 and their variance is σ², then 8σ² is equal to :

Solution

Maths sir · 4 steps

Use the two conditions to pin down a and d, write out all six terms, and the variance follows quickly.

1. Two equations. Take the first term a and common difference d. Then a₁ + a₃ = 2a + 2d = 10, so a + d = 5. The sum of six terms is 6a + 15d, and mean 19/2 gives 6a + 15d = 57, so 2a + 5d = 19.
2. Solve them. Put a = 5 - d into 2a + 5d = 19: 10 + 3d = 19, so d = 3 and a = 2. The numbers are 2, 5, 8, 11, 14, 17.

   σ² = (Σ xᵢ²)/6 - x̄² = 699/6 - (19/2)² = 233/2 - 361/4 = 105/4

3. Check the squares. 4 + 25 + 64 + 121 + 196 + 289 = 699. The AP shortcut agrees: σ² = d²(n²-1)/12 = (9 × 35)/12 = 105/4.
4. So. 8σ² = 8 × 105/4 = 210, which is option (b).

The idea to keep. For n terms in A.P. with common difference d, the variance is d²(n²-1)/12. It does not depend on the first term, so once you have d you are almost done.

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