X1-x2-x34 are numbers such that xi - xi-1 - 150 for all i - 1-2-3-9 and xi-1 - xi-2 - 0 for all i - 10-11-12-33 - Then median of x1- x2- - -x34 is

There are 34 terms. So mean of 17^th and 18^th term is the median. x_10=150 x_i+1 x_i = 2 ∀ i ∈{ 10,11,12,…,33 } ⇒ x_10, x_11, …,x_34 represents an A.P. wi ...

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Standard XII

Mathematics

Sum of n Terms

x1,x2,…,x34 a...

Question

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

150

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140

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135

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137

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Solution

The correct option is C $135$
There are $34$ terms.
So mean of $17^{th}$ and $18^{th}$ term is the median.
$x_{10} = 150$
$x_{i + 1} - x_{i} = - 2 \forall i \in {10, 11, 12, \dots, 33}$
$\Rightarrow x_{10}, x_{11}, \dots, x_{34}$ represents an A.P. with common difference of $- 2$
$x_{17} = 150 + (8 - 1) (- 2) = 136$
$x_{18} = 150 + (9 - 1) (- 2) = 134$

Hence, median $= \frac{136 + 134}{2} = 135$

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x1,x2,…,x34 are numbers such that xi=xi+1=150 for all i∈{1,2,3,…,9} and xi+1−xi+2=0 for all i∈{10,11,12,…,33}. Then median of x1,x2,…,x34 is

The correct option is C 135There are 34 terms. So mean of 17th and 18th term is the median. x10=150 xi+1−xi=−2 ∀ i∈{10,11,12,…,33} ⇒x10,x11,…,x34 represents an A.P. with common difference of −2 x17=150+(8−1)(−2)=136 x18=150+(9−1)(−2)=134 Hence, median =136+1342=135

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