x_1, x_2, dots, x_{34} are numbers such that | vedclass

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(C) The sequence has $34$ terms. The median is the average of the $17^{	ext{th}}$ and $18^{	ext{th}}$ terms.Given $x_1 = x_2 = \dots = x_{10} = 150$

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$135$

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(C) The sequence has $34$ terms. The median is the average of the $17^{	ext{th}}$ and $18^{	ext{th}}$ terms.Given $x_1 = x_2 = \dots = x_{10} = 150$.For $i \ge 10$,the sequence follows $x_{i+1} = x_i - 2$,which is an arithmetic progression with first term $a = x_{10} = 150$ and common difference $d = -2$.The $n^{	ext{th}}$ term of this progression (starting from $i=10$) is $x_{10+k} = 150 + k(-2)$.For the $17^{	ext{th}}$ term,$10+k = 17 \implies k = 7$,so $x_{17} = 150 + 7(-2) = 150 - 14 = 136$.For the $18^{	ext{th}}$ term,$10+k = 18 \implies k = 8$,so $x_{18} = 150 + 8(-2) = 150 - 16 = 134$.The median is $\frac{x_{17} + x_{18}}{2} = \frac{136 + 134}{2} = \frac{270}{2} = 135$.

$x_1, x_2, \dots, x_{34}$ are numbers such that $x_i = 150$ for all $i \in \{1, 2, \dots, 10\}$ and $x_{i+1} - x_i = -2$ for all $i \in \{10, 11, \dots, 33\}$. Find the median of $x_1, x_2, \dots, x_{34}$.

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