Question

Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$, be in AP 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. 105
• B. 210
• C. 200
• D. 220

Answer 1

The Correct Option is B

To solve this problem, we need to understand the given information about the sequence of numbers. We know that the six numbers \(a_1, a_2, a_3, a_4, a_5, a_6\) are in an arithmetic progression (AP), and we have two conditions:

1. \(a_1 + a_3 = 10\)
2. The mean of the six numbers is \(\frac{19}{2}\)

Step 1: Expressing the Terms of the Arithmetic Progression

Let the first term be \(a_1 = a\) and the common difference be \(d\). The terms of the AP can be expressed as:

• \(a_1 = a\)
• \(a_2 = a + d\)
• \(a_3 = a + 2d\)
• \(a_4 = a + 3d\)
• \(a_5 = a + 4d\)
• \(a_6 = a + 5d\)

Step 2: Using the Given Information

From the problem, we have:

• \(a + (a + 2d) = 10\)

Solving this gives:

\(2a + 2d = 10 \Rightarrow a + d = 5\)

The mean of the six numbers is:

\(\frac{a + (a+d) + (a+2d) + (a+3d) + (a+4d) + (a+5d)}{6} = \frac{19}{2}\)

Simplifying, we get:

\(\frac{6a + 15d}{6} = \frac{19}{2} \Rightarrow 6a + 15d = 57\)

We solve the equations:

• \(a + d = 5 \ldots (1)\)
• \(6a + 15d = 57 \ldots (2)\)

Substitute equation (1) into equation (2):

Let \(a = 5 - d\), substitute in equation (2):

\(6(5-d) + 15d = 57 \Rightarrow 30 - 6d + 15d = 57\)

Solving, we find:

\(9d = 27 \Rightarrow d = 3\)

Substituting \(d = 3\) into \(a + d = 5\):

\(a = 5 - 3 = 2\)

Step 3: Calculating the Variance

Variance of an AP sequence \(\sigma^2\) can be calculated using:

\(\sigma^2 = \frac{(n^2 - 1)}{12} \cdot d^2\)

Here, \(n=6\) (number of terms), and \(d=3\), thus:

\(\sigma^2 = \frac{(36 - 1)}{12} \cdot 9 = \frac{35}{12} \cdot 9 = \frac{315}{12} = \frac{105}{4}\)

The value of \(8 \sigma^2\) is:

\(8 \cdot \frac{105}{4} = 210\)

Therefore, the correct answer is 210.