If the first term of a G.P. `a_(1),a_(2),a_(3)` is unity such that `4a_(2)+5a_(3)` is least, then the common ratio of the G.P. is

Options

A. `2/5`
B. `-2/5`
C. `-3/5`
D. `-1/5`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript:

### Step 1: Identify the terms of the G.P.
Given that the first term of the G.P. is unity, we have:
- \( a_1 = 1 \)
- Let the common ratio be \( r \).
- Therefore, the second term \( a_2 = r \) and the third term \( a_3 = r^2 \).

### Step 2: Formulate the expression to minimize
We need to minimize the expression:
\[ 4a_2 + 5a_3 = 4r + 5r^2 \]

### Step 3: Recognize the form of the expression
The expression \( 4r + 5r^2 \) is a quadratic function in terms of \( r \). It can be rewritten as:
\[ f(r) = 5r^2 + 4r \]
where \( a = 5 \) and \( b = 4 \).

### Step 4: Find the vertex of the quadratic
The minimum value of a quadratic function \( ax^2 + bx + c \) occurs at:
\[ r = -\frac{b}{2a} \]
Substituting the values of \( a \) and \( b \):
\[ r = -\frac{4}{2 \cdot 5} = -\frac{4}{10} = -\frac{2}{5} \]

### Step 5: Conclusion
Thus, the common ratio \( r \) that minimizes \( 4a_2 + 5a_3 \) is:
\[ \boxed{-\frac{2}{5}} \]

---