Let 9=x_1

Question

Let $9=x_1<x_2<\cdots<x_7$ be in an A.P. with common difference d. If the standard deviation of $x_1,x_2,\ldots,x_7$ is 4 and the mean is $\bar{x}$, then $\bar{x}+x_6$ is equal to:

Accepted Answer

Terms: $9, 9+d, 9+2d,\ldots,9+6d$, i.e. $9+kd$ for $k=0,\ldots,6$. Shift by subtracting 9: $0,d,2d,\ldots,6d$.

Mean of shifted set $= \dfrac{21d}{7}=3d$, so $\bar x = 9+3d$.

$$16 = \frac{1}{7}(0^2+1^2+\cdots+6^2)d^2 - 9d^2 = \frac{91}{7}d^2-9d^2 = 13d^2-9d^2=4d^2$$

$$d^2=4 \implies d=2 \ (\text{increasing A.P.})$$

$\bar x = 9+3(2)=15$; $x_6 = 9+5(2)=19$.

$$\bar x+x_6 = 15+19=34$$

Solutions · 1