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 ${σ}^2$, then $8{σ}^2$ is equal to :

• A. 105
• B. 210
• C. 200
• D. 220

Answer 1

Answer: (B)

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