Let `a_(1), a_(2), a_(3)`, … be a G. P. such that `a_(1)lt0`, `a_(1)+a_(2)=4` and `a_(3)+a_(4)=16`. If `sum_(i=1)^(9) a_(i) = 4lambda`, then `lambda` is equal to :

Options

A. `(511)/(3)`
B. 171
C. `-513`
D. `-171`

Answer

To solve the problem step by step, we will analyze the given information about the geometric progression (G.P.) and derive the required value of \( \lambda \).

### Step 1: Define the terms of the G.P.
Let the first term of the G.P. be \( a_1 = a \) and the common ratio be \( r \). The terms of the G.P. can be expressed as:
- \( a_1 = a \)
- \( a_2 = ar \)
- \( a_3 = ar^2 \)
- \( a_4 = ar^3 \)

### Step 2: Set up the equations based on the problem statement
We know:
1. \( a_1 + a_2 = 4 \)
\[ a + ar = 4 \quad \text{(Equation 1)} \]
Factoring out \( a \):
\[ a(1 + r) = 4 \]

2. \( a_3 + a_4 = 16 \)
\[ ar^2 + ar^3 = 16 \quad \text{(Equation 2)} \]
Factoring out \( ar^2 \):
\[ ar^2(1 + r) = 16 \]

### Step 3: Divide the two equations
Dividing Equation 2 by Equation 1:
\[ \frac{ar^2(1 + r)}{a(1 + r)} = \frac{16}{4} \]
This simplifies to:
\[ r^2 = 4 \]
Taking the square root gives:
\[ r = 2 \quad \text{or} \quad r = -2 \]

### Step 4: Analyze the values of \( r \)
Since \( a_1 < 0 \), we will check both cases for \( r \).

1. **If \( r = 2 \)**:
Substituting into Equation 1:
\[ a(1 + 2) = 4 \Rightarrow 3a = 4 \Rightarrow a = \frac{4}{3} \]
This value of \( a \) is not acceptable since \( a_1 \) must be less than 0.

2. **If \( r = -2 \)**:
Substituting into Equation 1:
\[ a(1 - 2) = 4 \Rightarrow -a = 4 \Rightarrow a = -4 \]
This value of \( a \) is acceptable since \( a_1 = -4 < 0 \).

### Step 5: Calculate the sum of the first 9 terms
The sum of the first \( n \) terms of a G.P. is given by:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
For \( n = 9 \):
\[ S_9 = \frac{-4(1 - (-2)^9)}{1 - (-2)} = \frac{-4(1 + 512)}{1 + 2} = \frac{-4 \times 513}{3} \]
Calculating this gives:
\[ S_9 = \frac{-2052}{3} = -684 \]

### Step 6: Relate the sum to \( \lambda \)
We know from the problem statement:
\[ S_9 = 4\lambda \]
Thus:
\[ -684 = 4\lambda \Rightarrow \lambda = \frac{-684}{4} = -171 \]

### Final Answer
\[ \lambda = -171 \]