Let `{a_n}` be a GP such that `a_4/a_6 =1 /4 and a_2 + a_5 = 216`. Then `a_1` is equal to

Options

A. `12 or 108/7`
B. `10`
C. `7 or 54/7`
D. None of these

Explanation

To solve the problem, we need to find the value of \( a_1 \) given the conditions of the geometric progression (GP).

### Step 1: Understand the terms of the GP
In a GP, 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.

### Step 2: Write the given conditions using the GP formula
From the problem, we have:
1. \( \frac{a_4}{a_6} = \frac{1}{4} \)
2. \( a_2 + a_5 = 216 \)

Using the formula for the terms of the GP:
- \( a_4 = a_1 \cdot r^3 \)
- \( a_6 = a_1 \cdot r^5 \)

Substituting these into the first condition:
\[ \frac{a_1 \cdot r^3}{a_1 \cdot r^5} = \frac{1}{4} \]
This simplifies to:
\[ \frac{r^3}{r^5} = \frac{1}{4} \]
\[ \frac{1}{r^2} = \frac{1}{4} \]
Thus, we have:
\[ r^2 = 4 \quad \Rightarrow \quad r = 2 \text{ or } r = -2 \]

### Step 3: Use the second condition
Now, using the second condition \( a_2 + a_5 = 216 \):
- \( a_2 = a_1 \cdot r \)
- \( a_5 = a_1 \cdot r^4 \)

Substituting these into the second condition:
\[ a_1 \cdot r + a_1 \cdot r^4 = 216 \]
Factoring out \( a_1 \):
\[ a_1 (r + r^4) = 216 \]

### Step 4: Calculate for both values of \( r \)

**Case 1: \( r = 2 \)**
\[ r + r^4 = 2 + 2^4 = 2 + 16 = 18 \]
Thus,
\[ a_1 \cdot 18 = 216 \quad \Rightarrow \quad a_1 = \frac{216}{18} = 12 \]

**Case 2: \( r = -2 \)**
\[ r + r^4 = -2 + (-2)^4 = -2 + 16 = 14 \]
Thus,
\[ a_1 \cdot 14 = 216 \quad \Rightarrow \quad a_1 = \frac{216}{14} = \frac{108}{7} \]

### Conclusion
The possible values for \( a_1 \) are \( 12 \) and \( \frac{108}{7} \).

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

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