If x _ 1 , x _ 2 , x _ 3 ... x _ 34 are numbers such that x _ i = x _ i + 1 = 150, i ∈ \ 1,2, 3,4,.... 9\ and x _ i + 1 - x _ i = - 2, i ∈ \ 10,11,.... 33\ , then median of x _ 1 , x _ 2 , x _ 3 ...... x _ 34 is

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Statistics

NTA Abhyas 2020 · Mathematics · Statistics

If x _ 1 , x _ 2 , x _ 3 ... x _ 34 are numbers such that x _ i = x...

{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}

Options

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

Correct answer

B. $135$

Step-by-step solution

Total no. of term

= 34

So mean of

{17}^{th}

and

{18}^{th}

term is median

{x}_{10 + n} = 136, {x}_{18} = 134

Hence, median

= 135

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