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

The correct answer is (C) : 210
$a_1+a_3=10=a_1+d\Rightarrow 5$
$a_1+a_2+a_3+a_4+a_5+a_6=57$
$\Rightarrow \frac{6}{2}\left[a_1+a_6\right]=57$
$\Rightarrow a_1+a_6=19$
$\Rightarrow 2a_1+5d=19\text{and}{a}_1+d=5$
$\Rightarrow a_1=2,d=3$
Numbers : $2,5,8,11,14,17$
Variance $={\sigma }^2=$ mean of squares-square of mean
$=\frac{2^2+5^2+8^2+(11{)}^2+(14{)}^2+(17{)}^2}{6}-{\left(\frac{19}{2}\right)}^2$
$=\frac{699}{6}-\frac{361}{4}=\frac{105}{4}$
$8{\sigma }^2=210$

Answer 2

Step 1: Using the given condition \( a_1 + a_3 = 10 \)

The general terms of an arithmetic progression are: 

\( a_1, a_1 + d, a_1 + 2d, a_1 + 3d, a_1 + 4d, a_1 + 5d \)

From the condition \( a_1 + a_3 = 10 \), we get:

\[ a_1 + (a_1 + 2d) = 10 \quad \Rightarrow \quad 2a_1 + 2d = 10 \quad \Rightarrow \quad a_1 + d = 5 \quad \cdots (1) \] 

Step 2: Using the mean of the numbers

The mean of the six numbers is:

\[ \text{Mean} = \frac{a_1 + (a_1 + d) + (a_1 + 2d) + (a_1 + 3d) + (a_1 + 4d) + (a_1 + 5d)}{6} \] Simplifying this expression: \[ \text{Mean} = \frac{6a_1 + 15d}{6} = a_1 + \frac{5d}{2} \] Given that the mean is \( \frac{19}{2} \), we have: \[ a_1 + \frac{5d}{2} = \frac{19}{2} \quad \Rightarrow \quad 2a_1 + 5d = 19 \quad \cdots (2) \] 

Step 3: Solving equations (1) and (2)

From equation (1), we know:

\[ a_1 = 5 - d \] Substituting this into equation (2): \[ 2(5 - d) + 5d = 19 \quad \Rightarrow \quad 10 - 2d + 5d = 19 \quad \Rightarrow \quad 3d = 9 \quad \Rightarrow \quad d = 3 \] From \( a_1 = 5 - d \), we get: \[ a_1 = 5 - 3 = 2 \] 

Step 4: Finding the six numbers

The six numbers in the arithmetic progression are:

\[ a_1 = 2, \quad a_2 = 5, \quad a_3 = 8, \quad a_4 = 11, \quad a_5 = 14, \quad a_6 = 17 \] 

Step 5: Calculating variance (\( \sigma^2 \))

The variance formula is:

\[ \sigma^2 = \text{mean of squares} - (\text{square of mean}) \] The mean of squares is: \[ \frac{2^2 + 5^2 + 8^2 + 11^2 + 14^2 + 17^2}{6} = \frac{4 + 25 + 64 + 121 + 196 + 289}{6} = \frac{699}{6} = 116.5 \] The square of the mean is: \[ \left( \frac{19}{2} \right)^2 = \frac{361}{4} = 90.25 \] Therefore, the variance is: \[ \sigma^2 = 116.5 - 90.25 = 26.25 \] 

Step 6: Calculating \( 8\sigma^2 \) The final result is:

 

\[ 8\sigma^2 = 8 \times 26.25 = 210 \]