In a G.P., `T_(2) + T_(5) = 216 and T_(4) : T_(6) = 1 : 4` and all terms are integers, then its first term is

Options

A. 16
B. 14
C. 12
D. none

Answer

To solve the problem step by step, we will use the properties of a geometric progression (G.P.).

### Step 1: Define the terms of G.P.
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). The general term of a G.P. is given by:
\[ T_n = a \cdot r^{n-1} \]
Thus, we can express the required terms as:
- \( T_2 = a \cdot r^{2-1} = a \cdot r \)
- \( T_5 = a \cdot r^{5-1} = a \cdot r^4 \)
- \( T_4 = a \cdot r^{3} \)
- \( T_6 = a \cdot r^{5} \)

### Step 2: Set up the equations
From the problem, we have two equations:
1. \( T_2 + T_5 = 216 \)
\[ a \cdot r + a \cdot r^4 = 216 \quad \text{(Equation 1)} \]
This can be factored as:
\[ a(r + r^4) = 216 \]

2. The ratio \( T_4 : T_6 = 1 : 4 \)
\[ \frac{T_4}{T_6} = \frac{1}{4} \implies \frac{a \cdot r^3}{a \cdot r^5} = \frac{1}{4} \]
Simplifying this gives:
\[ \frac{r^3}{r^5} = \frac{1}{4} \implies \frac{1}{r^2} = \frac{1}{4} \implies r^2 = 4 \implies r = 2 \text{ or } r = -2 \]

### Step 3: Substitute \( r \) values into Equation 1
#### Case 1: \( r = 2 \)
Substituting \( r = 2 \) into Equation 1:
\[ a(2 + 2^4) = 216 \]
Calculating \( 2^4 \):
\[ a(2 + 16) = 216 \implies a \cdot 18 = 216 \implies a = \frac{216}{18} = 12 \]

#### Case 2: \( r = -2 \)
Substituting \( r = -2 \) into Equation 1:
\[ a(-2 + (-2)^4) = 216 \]
Calculating \( (-2)^4 \):
\[ a(-2 + 16) = 216 \implies a \cdot 14 = 216 \implies a = \frac{216}{14} \approx 15.43 \quad \text{(not an integer)} \]

### Conclusion
Since \( a \) must be an integer, we reject the case where \( r = -2 \). Therefore, the first term \( a \) of the G.P. is:
\[ \boxed{12} \]