Verified Solution • 2025–26 Session moderate

QJEEM23Ma1888: JEE Main 2023 Mathematics - Class 11 JEEM23Ma1888 Solution: Let be in an A.P. with common difference d. If the standard…

Let 9=x_{1} < x_{2} < \ldots < x_{7} be in an A.P. with common difference d. If the standard deviation of x1,x2...,x7x_{1}, x_{2}..., x_{7} is 4 and the mean is xˉ\bar{x} , then xˉ+x6\bar{x}+x_{6} is equal to :

Multiple Choice Options

  • A 2(9+87)2\left(9+\frac{8}{\sqrt{7}}\right)
  • B 25
  • C 18(1+13)18\left(1+\frac{1}{\sqrt{3}}\right)
  • D 34

Step-by-Step Verified Solution

amp; Mean ⇒xˉ=∑i=17xi7=72[2a+6d]7=a+3d=x4amp; Variance =∑i=17(xi−xˉ)27=(4)2⇒∑i=17(xi−x4)27=16amp;⇒(3d)2+(2d)2+d2+0+d2+(2d)2+(3d)27=16amp;=4d2=16⇒d=2amp;⇒xˉ=9+3(2)=15amp;x6=a+5d=9+5(2)=19⇒xˉ+x6=34\begin{aligned} & \text { Mean } \Rightarrow \bar{x}=\frac{\sum\limits_{i=1}^7 x_i}{7}=\frac{\frac{7}{2}[2 a+6 d]}{7}=a+3 d=x_4 \\\\ & \text { Variance }=\frac{\sum\limits_{i=1}^7\left(x_i-\bar{x}\right)^2}{7}=(4)^2 \Rightarrow \frac{\sum\limits_{i=1}^7\left(x_i-x_4\right)^2}{7}=16 \\\\ & \Rightarrow \frac{(3 d)^2+(2 d)^2+d^2+0+d^2+(2 d)^2+(3 d)^2}{7}=16 \\\\ & =4 d^2=16 \Rightarrow d=2 \\\\ & \Rightarrow \bar{x}=9+3(2)=15 \\\\ & x_6=a+5 d=9+5(2)=19 \Rightarrow \bar{x}+x_6=34\end{aligned}

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