In a GP, first term is 1. If `4T_(2) + 5T_(3)` is minimum, then its common ratio is

Options

A. `(2)/(5)`
B. `-(2)/(5)`
C. `(3)/(5)`
D. `-(3)/(5)`

Answer

To solve the problem, we need to find the common ratio \( r \) of a geometric progression (GP) where the first term \( T_1 = 1 \) and the expression \( 4T_2 + 5T_3 \) is minimized.

### Step-by-step Solution:

1. **Identify the Terms of the GP**:
- The first term \( T_1 = 1 \).
- The second term \( T_2 = T_1 \cdot r = 1 \cdot r = r \).
- The third term \( T_3 = T_1 \cdot r^2 = 1 \cdot r^2 = r^2 \).

2. **Set Up the Expression**:
- We need to minimize the expression \( 4T_2 + 5T_3 \).
- Substitute \( T_2 \) and \( T_3 \):
\[ S = 4T_2 + 5T_3 = 4r + 5r^2. \]

3. **Differentiate the Expression**:
- To find the minimum value, we differentiate \( S \) with respect to \( r \):
\[ \frac{dS}{dr} = 4 + 10r. \]

4. **Set the Derivative to Zero**:
- Set the derivative equal to zero to find critical points:
\[ 4 + 10r = 0. \]
- Solve for \( r \):
\[ 10r = -4 \implies r = -\frac{2}{5}. \]

5. **Conclusion**:
- The common ratio \( r \) that minimizes \( 4T_2 + 5T_3 \) is:
\[ r = -\frac{2}{5}. \]

### Final Answer:
The common ratio is \( r = -\frac{2}{5} \).

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