Question

If $x_{1},x_{2},x_{3}...x_{34}$ are numbers such that $x_{i}=x_{i + 1}=150,\forall i\in \left\{1,2 , 3,4 , . . . . 9\right\}$ and $x_{i + 1}-x_{i}=-2,\forall i\in \left\{10,11 , . . . . 33\right\}$ , then median of $x_{1},x_{2},x_{3}......x_{34}$ is

• A. $134$
• B. $135$
• C. $148$
• D. $150$

Answer 1

Answer: (B)

Total no. of term $=34$
So mean of $17^{t h}$ and $18^{t h}$ term is median
$x_{10 + n}=136,x_{18}=134$
Hence, median $=135$