Question

$x_1,x_2,\dots ,x_{34}$ are numbers such that $x_i=x_{i+1}=150\forall i\in \{1,2,3,\dots ,9\}$ and $x_{i+1}-x_i+2=0\forall i\in \{10,11,12,\dots ,33\}.$ Then median of $x_1,x_2,\dots ,x_{34}$ is

• A. $133$
• B. $140$
• C. $135$
• D. $137$

Answer 1

The correct option is C $135$

There are total $34$ terms.
So mean of ${17}^{\text{th}}$ and ${18}^{\text{th}}$ term is the median.
Now, $x_{10}=150$
and $x_{i+1}-x_i=-2\forall i\in \{10,11,12,\dots ,33\}$
$\Rightarrow x_{10},x_{11},\dots ,x_{34}$ are in A.P. with first term and common difference equal to $150$ and $-2$ respectively.
$\therefore x_{17}=150+(8-1)(-2)=136$
and $x_{18}=150+(9-1)(-2)=134$
Hence, median $=\frac{x_{17}+x_{18}}{2}=135$