If x1,x2,x3...x34 are numbers such that xi=xi + 1=150,∀ i∈ 1,2 , 3,4 , . . . . 9 and xi + 1-xi=-2,∀ i∈ 10,11 , . . . . 33 , then median of x1,x2,x3......x34 is

Question

If x1,x2,x3...x34 are numbers such that xi=xi + 1=150,∀ i∈ 1,2 , 3,4 , . . . . 9 and xi + 1-xi=-2,∀ i∈ 10,11 , . . . . 33 , then median of x1,x2,x3......x34 is (A) 134 (B) 135 (C) 148 (D) 150

Tardigrade Admin · Accepted Answer

Total no. of term =34 So mean of 17t h and 18t h term is median x10 + n=136,x18=134 Hence, median =135

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

$134$

$135$

$148$

$150$

Total no. of term $= 34$
So mean of $1 7^{t h}$ and $1 8^{t h}$ term is median
$x_{10 + n} = 136, x_{18} = 134$
Hence, median $= 135$

Q. If x1​,x2​,x3​...x34​ are numbers such that xi​=xi+1​=150,∀i∈{1,2,3,4,....9} and xi+1​−xi​=−2,∀i∈{10,11,....33} , then median of x1​,x2​,x3​......x34​ is

134

135

148

150