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)

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