If the mean of a set of observations `x_(1), x_(2),……,x_(10)` is 40, then the mean of `x_(1)+4, x_(2)+8, x_(3)+12,…….,x_(10)+40` is

Options

A. 54
B. 62
C. 38
D. 50

Answer

To find the mean of the new set of observations \( x_1 + 4, x_2 + 8, x_3 + 12, \ldots, x_{10} + 40 \), we can follow these steps:

### Step 1: Calculate the sum of the original observations
Given that the mean of the observations \( x_1, x_2, \ldots, x_{10} \) is 40, we can express this mathematically:
\[ \text{Mean} = \frac{x_1 + x_2 + x_3 + \ldots + x_{10}}{10} = 40 \]
From this, we can find the sum of the observations:
\[ x_1 + x_2 + x_3 + \ldots + x_{10} = 40 \times 10 = 400 \]

### Step 2: Write the new observations
The new observations are:
\[ x_1 + 4, x_2 + 8, x_3 + 12, \ldots, x_{10} + 40 \]
We can express this as:
\[ (x_1 + 4) + (x_2 + 8) + (x_3 + 12) + \ldots + (x_{10} + 40) \]

### Step 3: Separate the sums
We can separate the sum of the new observations into two parts:
\[ \text{Sum of new observations} = (x_1 + x_2 + x_3 + \ldots + x_{10}) + (4 + 8 + 12 + \ldots + 40) \]
We already know that:
\[ x_1 + x_2 + x_3 + \ldots + x_{10} = 400 \]

### Step 4: Calculate the sum of the added constants
Next, we need to calculate the sum \( 4 + 8 + 12 + \ldots + 40 \). This is an arithmetic series where:
- The first term \( a = 4 \)
- The last term \( l = 40 \)
- The common difference \( d = 4 \)

To find the number of terms \( n \):
\[ n = \frac{l - a}{d} + 1 = \frac{40 - 4}{4} + 1 = 10 \]

Now, we can calculate the sum of the series:
\[ \text{Sum} = \frac{n}{2} \times (a + l) = \frac{10}{2} \times (4 + 40) = 5 \times 44 = 220 \]

### Step 5: Combine the sums
Now we can combine the sums:
\[ \text{Total sum of new observations} = 400 + 220 = 620 \]

### Step 6: Calculate the mean of the new observations
Finally, we calculate the mean of the new observations:
\[ \text{Mean} = \frac{\text{Total sum}}{\text{Number of observations}} = \frac{620}{10} = 62 \]

### Final Answer
Thus, the mean of the new set of observations is \( \boxed{62} \).