If `x_(1), x_(2), x_(3)…..x_(34)` are numbers such that `x_(i)=x_(i+1)=150, AA I in {1,2, 3, 4, …..9}` and `x_(i+1)-x_(i)=-2, AA I in {10, 11, ……..33}`, then median of `x_(1), x_(2),x_(3)………x_(34)` is

Options

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

Answer

To find the median of the sequence \( x_1, x_2, x_3, \ldots, x_{34} \), we follow these steps:

### Step 1: Identify the first 10 terms
Given that \( x_i = 150 \) for \( i = 1, 2, 3, \ldots, 9 \) and \( x_{10} = 150 \), we have:
\[ x_1 = x_2 = x_3 = \ldots = x_{10} = 150 \]

### Step 2: Determine the next terms
For \( i = 10, 11, \ldots, 33 \), the relationship is given by \( x_{i+1} - x_i = -2 \). This means:
\[ x_{i+1} = x_i - 2 \]
Starting from \( x_{10} = 150 \):
- \( x_{11} = x_{10} - 2 = 150 - 2 = 148 \)
- \( x_{12} = x_{11} - 2 = 148 - 2 = 146 \)
- \( x_{13} = x_{12} - 2 = 146 - 2 = 144 \)
- Continuing this pattern, we can find the subsequent terms.

### Step 3: Calculate the terms from \( x_{11} \) to \( x_{34} \)
Using the formula \( x_{i} = 150 - 2(i - 10) \) for \( i = 11, 12, \ldots, 34 \):
- \( x_{11} = 150 - 2(11 - 10) = 150 - 2 = 148 \)
- \( x_{12} = 150 - 2(12 - 10) = 150 - 4 = 146 \)
- \( x_{13} = 150 - 2(13 - 10) = 150 - 6 = 144 \)
- Continuing this way, we find:
- \( x_{14} = 142 \)
- \( x_{15} = 140 \)
- \( x_{16} = 138 \)
- \( x_{17} = 136 \)
- \( x_{18} = 134 \)
- \( x_{19} = 132 \)
- \( x_{20} = 130 \)
- \( x_{21} = 128 \)
- \( x_{22} = 126 \)
- \( x_{23} = 124 \)
- \( x_{24} = 122 \)
- \( x_{25} = 120 \)
- \( x_{26} = 118 \)
- \( x_{27} = 116 \)
- \( x_{28} = 114 \)
- \( x_{29} = 112 \)
- \( x_{30} = 110 \)
- \( x_{31} = 108 \)
- \( x_{32} = 106 \)
- \( x_{33} = 104 \)
- \( x_{34} = 102 \)

### Step 4: Find the 17th and 18th terms
Now we need to find the 17th and 18th terms:
- \( x_{17} = 136 \)
- \( x_{18} = 134 \)

### Step 5: Calculate the median
The median \( M \) is given by:
\[ M = \frac{x_{17} + x_{18}}{2} = \frac{136 + 134}{2} = \frac{270}{2} = 135 \]

### Final Answer
The median of \( x_1, x_2, \ldots, x_{34} \) is \( \boxed{135} \).