If variance of `1, 3, 5, 7, …….. 215` is `A^(2)` then standard deviation of `1, 5, 9, 13, …..429` in terms of `A` is :

Options

A. `2A`
B. `2A^(2)`
C. `|A|`
D. `2|A|`

Explanation

To solve the problem, we need to find the standard deviation of the series \(1, 5, 9, 13, \ldots, 429\) in terms of the variance \(A^2\) of the series \(1, 3, 5, 7, \ldots, 215\).

### Step-by-Step Solution:

1. **Identify the Series:**
- The first series is \(1, 3, 5, 7, \ldots, 215\). This is an arithmetic progression (AP) with the first term \(a_1 = 1\) and common difference \(d_1 = 2\).
- The second series is \(1, 5, 9, 13, \ldots, 429\). This is also an AP with the first term \(a_2 = 1\) and common difference \(d_2 = 4\).

2. **Find the Number of Terms:**
- For the first series, the last term is \(215\). The number of terms \(n_1\) can be found using the formula for the \(n\)-th term of an AP:
\[ a_n = a_1 + (n-1) \cdot d_1 \]
Setting \(a_n = 215\):
\[ 215 = 1 + (n_1 - 1) \cdot 2 \implies 214 = (n_1 - 1) \cdot 2 \implies n_1 - 1 = 107 \implies n_1 = 108 \]

- For the second series, the last term is \(429\):
\[ 429 = 1 + (n_2 - 1) \cdot 4 \implies 428 = (n_2 - 1) \cdot 4 \implies n_2 - 1 = 107 \implies n_2 = 108 \]

3. **Calculate the Variance of the First Series:**
- The mean \(\mu_1\) of the first series can be calculated as:
\[ \mu_1 = \frac{\text{Sum of terms}}{n_1} = \frac{n_1}{2} \cdot (a_1 + a_n) = \frac{108}{2} \cdot (1 + 215) = 54 \cdot 216 = 11664 \]
- The variance \(A^2\) is given by:
\[ A^2 = \frac{1}{n_1} \sum (x_i - \mu_1)^2 \]
- For an AP, the variance can also be calculated using:
\[ \text{Variance} = \frac{(d_1^2)(n_1^2 - 1)}{12} \]
Substituting \(d_1 = 2\) and \(n_1 = 108\):
\[ A^2 = \frac{(2^2)(108^2 - 1)}{12} = \frac{4 \cdot (11664 - 1)}{12} = \frac{4 \cdot 11663}{12} \]

4. **Calculate the Standard Deviation of the Second Series:**
- The standard deviation \(\sigma_2\) of the second series is given by:
\[ \sigma_2 = \sqrt{\text{Variance of the second series}} \]
- The variance of the second series can be calculated similarly:
\[ \text{Variance} = \frac{(d_2^2)(n_2^2 - 1)}{12} \]
Substituting \(d_2 = 4\) and \(n_2 = 108\):
\[ \text{Variance} = \frac{(4^2)(108^2 - 1)}{12} = \frac{16 \cdot (11664 - 1)}{12} = \frac{16 \cdot 11663}{12} \]

5. **Relate the Two Variances:**
- The standard deviation of the second series can be expressed in terms of \(A\):
\[ \sigma_2 = \sqrt{4 \cdot A^2} = 2A \]

### Final Answer:
Thus, the standard deviation of the series \(1, 5, 9, 13, \ldots, 429\) in terms of \(A\) is:
\[ \boxed{2A} \]