Let `V_(1)=` variance of `{13, 16, 19…………103} and V_(2)=" variance of "{20, 26, 32…………..200}`. Then `V_(1):V_(2)` is

Options

A. `1:2`
B. `1:1`
C. `4:9`
D. `1:4`

Answer

To find the ratio of the variances \( V_1 : V_2 \) for the given sequences, we will follow these steps:

### Step 1: Identify the sequences
The first sequence is \( S_1 = \{ 13, 16, 19, \ldots, 103 \} \) and the second sequence is \( S_2 = \{ 20, 26, 32, \ldots, 200 \} \).

### Step 2: Determine the number of terms in each sequence
For the first sequence \( S_1 \):
- The first term \( a_1 = 13 \)
- The common difference \( d_1 = 3 \)
- The last term \( a_n = 103 \)

Using the formula for the \( n \)-th term of an arithmetic progression (AP):
\[ a_n = a_1 + (n - 1) d \]
Setting \( a_n = 103 \):
\[ 103 = 13 + (n - 1) \cdot 3 \]
\[ 103 - 13 = (n - 1) \cdot 3 \]
\[ 90 = (n - 1) \cdot 3 \]
\[ n - 1 = 30 \implies n = 31 \]
So, there are 31 terms in \( S_1 \).

For the second sequence \( S_2 \):
- The first term \( a_2 = 20 \)
- The common difference \( d_2 = 6 \)
- The last term \( a_n = 200 \)

Using the same formula:
\[ 200 = 20 + (n - 1) \cdot 6 \]
\[ 200 - 20 = (n - 1) \cdot 6 \]
\[ 180 = (n - 1) \cdot 6 \]
\[ n - 1 = 30 \implies n = 31 \]
So, there are also 31 terms in \( S_2 \).

### Step 3: Calculate the variance of each sequence
The variance \( V \) of a sequence is given by:
\[ V = \frac{1}{n} \sum (x_i - \bar{x})^2 \]
However, we can also use the property that variance is proportional to the square of the common difference when the number of terms is the same.

For \( S_1 \):
- The common difference \( d_1 = 3 \)
- Therefore, \( V_1 \propto d_1^2 = 3^2 = 9 \)

For \( S_2 \):
- The common difference \( d_2 = 6 \)
- Therefore, \( V_2 \propto d_2^2 = 6^2 = 36 \)

### Step 4: Find the ratio of variances
Now, we can find the ratio of the variances:
\[ \frac{V_1}{V_2} = \frac{9}{36} = \frac{1}{4} \]

### Final Answer
Thus, the ratio \( V_1 : V_2 \) is \( 1 : 4 \).

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