Let 9

Question

(B) The terms are $x_1, x_2, \ldots, x_7$ in $A.P.$ with $x_1 = 9$ and common difference $d > 0$.These can be written as $9, 9+d, 9+2d, \ldots, 9+6d$.

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(B) The terms are $x_1, x_2, \ldots, x_7$ in $A.P.$ with $x_1 = 9$ and common difference $d > 0$.These can be written as $9, 9+d, 9+2d, \ldots, 9+6d$.The mean $\overline{x} = \frac{1}{7} \sum_{i=0}^{6} (9+id) = 9 + \frac{d}{7} 	imes \frac{6 	imes 7}{2} = 9 + 3d$.The variance $\sigma^2$ is given by $\frac{1}{n} \sum x_i^2 - (\overline{x})^2$.Since standard deviation $\sigma = 4$,$\sigma^2 = 16$.Using the property of variance for an $A.P.$,$\sigma^2 = d^2 \frac{n^2-1}{12}$.Here $n=7$,so $\sigma^2 = d^2 \frac{49-1}{12} = d^2 \frac{48}{12} = 4d^2$.Given $4d^2 = 16$,we get $d^2 = 4$,so $d = 2$.Now,$\overline{x} = 9 + 3(2) = 15$.$x_6 = 9 + 5d = 9 + 5(2) = 19$.Therefore,$\overline{x} + x_6 = 15 + 19 = 34$.

Let $9 < x_1 < x_2 < \ldots < x_7$ 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 $\overline{x}$,then $\overline{x} + x_6$ is equal to:

Similar Questions

The common difference of the $A.P.: a_{1}, a_{2}, ..., a_{m}$ is $13$ more than the common difference of the $A.P.: b_{1}, b_{2}, ..., b_{n}$. If $b_{31} = -277$,$b_{43} = -385$ and $a_{78} = 327$,then $a_{1}$ is equal to

The minimum value of ${\left( {\frac{3}{a} - 1} \right)^2} + {\left( {\frac{a}{b} - 1} \right)^2} + {\left( {\frac{b}{c} - 1} \right)^2} + {\left( {3c - 1} \right)^2}$ where $0 < a, b, c \leqslant 9$,is $p - q\sqrt{r}$; $p, q, r \in I$ and $q, r$ are coprime,then $(p + q + r)$ is equal to

There are $n$ arithmetic means between $7$ and $71$. If the $5^{th}$ arithmetic mean is $27$,then $n = ......$

If the $p^{th}$ term of an arithmetic progression is $q$ and its $q^{th}$ term is $p$,then what is its $(p + q)^{th}$ term?

What is the sum of $n$ arithmetic means between $a$ and $b$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module