Let `a_1, a_2, a_3, ?a_10 `are in G.P. if `a_3/a_1 =25` then `a_9/a_5` is equal to
(A) `5^4`
(B) `4.5^4`
(C) `4.5^3`
(D) `5^3`

Options

A. `2(5^(2))`
B. `4(5^(2))`
C. `5^(4)`
D. `5^(3)`

Answer

To solve the problem, we need to find the value of \( \frac{a_9}{a_5} \) given that \( \frac{a_3}{a_1} = 25 \) and the terms \( a_1, a_2, a_3, \ldots, a_{10} \) are in a geometric progression (G.P.).

### Step-by-Step Solution:

1. **Understanding the terms in G.P.**:
In a G.P., the \( n \)-th term can be expressed as:
\[ a_n = a_1 \cdot r^{n-1} \]
where \( a_1 \) is the first term and \( r \) is the common ratio.

2. **Expressing \( a_3 \) and \( a_1 \)**:
We can express \( a_3 \) and \( a_1 \) using the formula for the \( n \)-th term:
\[ a_3 = a_1 \cdot r^{3-1} = a_1 \cdot r^2 \]
\[ a_1 = a_1 \cdot r^{1-1} = a_1 \]

3. **Setting up the ratio**:
Given that \( \frac{a_3}{a_1} = 25 \), we can substitute the expressions we found:
\[ \frac{a_1 \cdot r^2}{a_1} = 25 \]
This simplifies to:
\[ r^2 = 25 \]

4. **Finding \( r \)**:
Taking the square root of both sides, we find:
\[ r = 5 \quad \text{(since we consider the positive root in G.P.)} \]

5. **Finding \( \frac{a_9}{a_5} \)**:
Now we need to find \( \frac{a_9}{a_5} \):
\[ a_9 = a_1 \cdot r^{9-1} = a_1 \cdot r^8 \]
\[ a_5 = a_1 \cdot r^{5-1} = a_1 \cdot r^4 \]
Therefore, the ratio \( \frac{a_9}{a_5} \) is:
\[ \frac{a_9}{a_5} = \frac{a_1 \cdot r^8}{a_1 \cdot r^4} = \frac{r^8}{r^4} = r^{8-4} = r^4 \]

6. **Calculating \( r^4 \)**:
Since we found \( r = 5 \):
\[ r^4 = 5^4 \]

7. **Conclusion**:
Thus, we have:
\[ \frac{a_9}{a_5} = 5^4 \]
The answer is \( \boxed{5^4} \).