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. 105
• B. 210
• C. 200
• D. 220

Answer 1

Answer: (B)

$ a_1+a_3=10=a_1+d \Rightarrow 5$
$ a _1+ a _2+a_3+ a _{ 4 }+ a _5+ a _6= 5 7 $
$ \Rightarrow \frac{6}{2}\left[a_1+a_6\right]=57 $
$ \Rightarrow a_1+a_6=19 $
$ \Rightarrow 2 a_1+5 d=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$