Let `X_(1), X_(2), X_(3)…….` are in arithmetic progression with a common difference equal to d which is a two digit natural number. `y_(1), y_(2), y_(3)………` are in geometric progression with common ratio equal to 16. Arithmetic mean of `X_(1), X_(2)……..X_(n)` is equal to the arithmetic mean of `y_(1), y_(2)......y_(n)` which is equal to 5. If the arithmetic mean of `X_(6), X_(7)....X_(n+5)` is equal to the arithmetic mean of `y_(P+1), y_(P+2)......y_(P+n)` then d is equal to

Answer

To solve the problem, we need to find the common difference \( d \) of the arithmetic progression \( X_1, X_2, \ldots \) and ensure that it is a two-digit natural number. The arithmetic mean of both the arithmetic progression and the geometric progression is given to be 5.

### Step-by-Step Solution:

1. **Define the Arithmetic Progression (AP)**:
- Let the first term of the AP be \( X_1 = a \).
- The terms of the AP can be expressed as \( X_n = a + (n-1)d \).
- The arithmetic mean of the first \( n \) terms of the AP is given by:
\[ \text{AM}_{AP} = \frac{X_1 + X_2 + \ldots + X_n}{n} = \frac{n}{2} \cdot \frac{2a + (n-1)d}{n} = \frac{2a + (n-1)d}{2} \]
- Given that this mean equals 5, we have:
\[ 2a + (n-1)d = 10 \quad \text{(Equation 1)} \]

2. **Define the Geometric Progression (GP)**:
- Let the first term of the GP be \( Y_1 = b \) and the common ratio be \( r = 16 \).
- The terms of the GP can be expressed as \( Y_n = b \cdot 16^{n-1} \).
- The arithmetic mean of the first \( n \) terms of the GP is given by:
\[ \text{AM}_{GP} = \frac{Y_1 + Y_2 + \ldots + Y_n}{n} = \frac{b(1 + 16 + 16^2 + \ldots + 16^{n-1})}{n} = \frac{b \cdot \frac{16^n - 1}{15}}{n} \]
- Given that this mean also equals 5, we have:
\[ \frac{b(16^n - 1)}{15n} = 5 \quad \Rightarrow \quad b(16^n - 1) = 75n \quad \text{(Equation 2)} \]

3. **Mean of Specific Terms**:
- We need to find the mean of \( X_6, X_7, \ldots, X_{n+5} \):
\[ \text{AM}_{X_6} = \frac{X_6 + X_7 + \ldots + X_{n+5}}{n} = \frac{n}{2} \cdot \frac{2X_6 + (n-1)d}{n} = \frac{2X_6 + (n-1)d}{2} \]
- The first term \( X_6 = a + 5d \), so:
\[ \text{AM}_{X_6} = \frac{2(a + 5d) + (n-1)d}{2} = \frac{2a + 10d + (n-1)d}{2} = \frac{2a + (n + 9)d}{2} \]

4. **Mean of Specific Terms in GP**:
- The mean of \( Y_{p+1}, Y_{p+2}, \ldots, Y_{p+n} \):
\[ \text{AM}_{Y_{p}} = \frac{Y_{p+1} + Y_{p+2} + \ldots + Y_{p+n}}{n} = \frac{b(16^p + 16^{p+1} + \ldots + 16^{p+n-1})}{n} = \frac{b \cdot 16^p \cdot \frac{16^n - 1}{15}}{n} \]
- Setting this equal to the previous mean gives us:
\[ \frac{2a + (n + 9)d}{2} = \frac{b \cdot 16^p (16^n - 1)}{15n} \]

5. **Equating and Solving**:
- From the equations derived, we can substitute the values and simplify to find \( d \).
- After simplification, we find:
\[ 5d = 16^p - 1 \]
- Since \( d \) is a two-digit number, we can test values of \( p \):
- For \( p = 1 \):
\[ d = \frac{16^1 - 1}{5} = \frac{15}{5} = 3 \quad \text{(not two-digit)} \]
- For \( p = 2 \):
\[ d = \frac{16^2 - 1}{5} = \frac{255}{5} = 51 \quad \text{(valid)} \]
- For \( p = 3 \):
\[ d = \frac{16^3 - 1}{5} = \frac{4095}{5} = 819 \quad \text{(not two-digit)} \]

Thus, the only valid two-digit solution is \( d = 51 \).

### Final Answer:
\[ d = 51 \]