Let `{a_n}` bc a G.P. such that `a_4/a_6=1/4` and `a_2+a_5=216`. Then `a_1=`

Options

A. `12or,(108)/(7)`
B. 10
C. `7or,(54)/(7)`
D. none of these

Explanation

To solve the problem, we need to find the first term \( a_1 \) of the geometric progression (G.P.) given the conditions \( \frac{a_4}{a_6} = \frac{1}{4} \) and \( a_2 + a_5 = 216 \).

### Step-by-Step Solution:

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

2. **Expressing \( a_4 \) and \( a_6 \)**:
Using the formula for the \( n \)-th term:
\[ a_4 = a r^{3} \quad \text{and} \quad a_6 = a r^{5} \]

3. **Setting up the ratio**:
Given \( \frac{a_4}{a_6} = \frac{1}{4} \):
\[ \frac{a r^{3}}{a r^{5}} = \frac{1}{4} \]
Simplifying this gives:
\[ \frac{r^{3}}{r^{5}} = \frac{1}{4} \implies \frac{1}{r^{2}} = \frac{1}{4} \]
Therefore:
\[ r^{2} = 4 \implies r = 2 \quad \text{or} \quad r = -2 \]

4. **Using the second condition**:
Now, we use the second condition \( a_2 + a_5 = 216 \):
\[ a_2 = a r^{1} \quad \text{and} \quad a_5 = a r^{4} \]
Thus:
\[ a r + a r^{4} = 216 \]
Factoring out \( a \):
\[ a (r + r^{4}) = 216 \]

5. **Calculating for \( r = 2 \)**:
Substituting \( r = 2 \):
\[ r + r^{4} = 2 + 2^{4} = 2 + 16 = 18 \]
Therefore:
\[ a \cdot 18 = 216 \implies a = \frac{216}{18} = 12 \]

6. **Calculating for \( r = -2 \)**:
Now substituting \( r = -2 \):
\[ r + r^{4} = -2 + (-2)^{4} = -2 + 16 = 14 \]
Thus:
\[ a \cdot 14 = 216 \implies a = \frac{216}{14} = \frac{108}{7} \]

7. **Finding \( a_1 \)**:
The first term \( a_1 \) is simply \( a \):
- For \( r = 2 \), \( a_1 = 12 \).
- For \( r = -2 \), \( a_1 = \frac{108}{7} \).

### Final Answer:
Thus, \( a_1 \) can be either \( 12 \) or \( \frac{108}{7} \).