Question

Let the six numbers \( a_1, a_2, a_3, a_4, a_5, a_6 \) be in A.P., and \( a_1 + a_3 = 10 \). If the mean of these six numbers is \( \frac{19}{2} \) and their variance is \( \sigma^2 \), then \( 8\sigma^2 \) is equal to:

• A. \(220 \)
• B. \(210 \)
• C. \(200 \)
• D. \(105 \)
• E.

Hint:

Convert given conditions into equations in \(a_1\) and \(d\), find actual terms, then apply variance formula carefully.

Answer 1

The Correct Option is B

Concept: In an Arithmetic Progression (A.P.), each term differs from the previous one by a constant value called the common difference \(d\). If the first term is \(a_1\), then the six terms are: \[ a_1, \; a_1+d, \; a_1+2d, \; a_1+3d, \; a_1+4d, \; a_1+5d \] We use:
• Mean: \( \mu = \frac{\text{Sum of terms}}{n} \)
• Variance: \( \sigma^2 = \frac{1}{n}\sum x_i^2 - \mu^2 \)

Step 1: Using the condition \( a_1 + a_3 = 10 \)
We know: \[ a_3 = a_1 + 2d \] So, \[ a_1 + a_3 = a_1 + (a_1 + 2d) = 2a_1 + 2d \] Given: \[ 2a_1 + 2d = 10 \] Divide by 2: \[ a_1 + d = 5 \quad \cdots (1) \]

Step 2: Using the mean condition
Mean is: \[ \mu = \frac{19}{2} \] Sum of 6 terms: \[ S_6 = \frac{6}{2}(2a_1 + 5d) = 3(2a_1 + 5d) \] Mean: \[ \mu = \frac{S_6}{6} = \frac{2a_1 + 5d}{2} \] So, \[ \frac{2a_1 + 5d}{2} = \frac{19}{2} \] \[ 2a_1 + 5d = 19 \quad \cdots (2) \]

Step 3: Solving equations
From (1): \[ a_1 = 5 - d \] Substitute into (2): \[ 2(5 - d) + 5d = 19 \] \[ 10 - 2d + 5d = 19 \] \[ 10 + 3d = 19 \] \[ 3d = 9 \Rightarrow d = 3 \] Then, \[ a_1 = 5 - 3 = 2 \]

Step 4: Writing the A.P.
\[ 2, \; 5, \; 8, \; 11, \; 14, \; 17 \]

Step 5: Verifying the mean
\[ \mu = \frac{2+5+8+11+14+17}{6} = \frac{57}{6} = \frac{19}{2} \]

Step 6: Finding variance
Squares: \[ 4,\; 25,\; 64,\; 121,\; 196,\; 289 \] Sum: \[ 699 \] Mean of squares: \[ \frac{699}{6} = 116.5 \] Square of mean: \[ \left(\frac{19}{2}\right)^2 = \frac{361}{4} = 90.25 \] \[ \sigma^2 = 116.5 - 90.25 = 26.25 \]

Step 7: Final answer
\[ 8\sigma^2 = 8 \times 26.25 = 210 \] \[ \boxed{210} \]