`a_1, a_2, a_3 …..a_9` are in GP where `a_1 lt 0, a_1 + a_2 = 4, a_3 + a_4 = 16`, if `sum_(i=1)^9 a_i = 4 lambda`then `lambda` is equal to

Options

A. `-513`
B. `-171`
C. `171`
D. `-171/3`

Explanation

To solve the problem, we need to find the value of \( \lambda \) given the conditions about the geometric progression (GP) of the terms \( a_1, a_2, a_3, \ldots, a_9 \).

### Step-by-Step Solution:

1. **Define the Terms of the GP:**
Let \( a_1 = a \) and the common ratio be \( r \). Then the terms can be expressed as:
\[ a_2 = ar, \quad a_3 = ar^2, \quad a_4 = ar^3, \quad \ldots, \quad a_9 = ar^8 \]

2. **Use the Given Conditions:**
From the problem, we have the following equations:
- \( a_1 + a_2 = 4 \)
- \( a_3 + a_4 = 16 \)

Substituting the expressions for \( a_1 \) and \( a_2 \):
\[ a + ar = 4 \quad \text{(Equation 1)} \]
For \( a_3 \) and \( a_4 \):
\[ ar^2 + ar^3 = 16 \quad \text{(Equation 2)} \]

3. **Simplify the Equations:**
From Equation 1:
\[ a(1 + r) = 4 \implies a = \frac{4}{1 + r} \]

Substitute \( a \) into Equation 2:
\[ \frac{4}{1 + r} r^2 + \frac{4}{1 + r} r^3 = 16 \]
Factoring out \( \frac{4r^2}{1 + r} \):
\[ \frac{4r^2(1 + r)}{1 + r} = 16 \implies 4r^2 = 16 \implies r^2 = 4 \implies r = 2 \text{ or } r = -2 \]

4. **Determine the Values of \( a \):**
If \( r = 2 \):
\[ a = \frac{4}{1 + 2} = \frac{4}{3} \]
If \( r = -2 \):
\[ a = \frac{4}{1 - 2} = -4 \]
Since \( a_1 < 0 \), we take \( a = -4 \) and \( r = -2 \).

5. **Calculate the Sum of the First 9 Terms:**
The sum of the first \( n \) terms of a GP is given by:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
For \( n = 9 \):
\[ S_9 = \frac{-4(1 - (-2)^9)}{1 - (-2)} = \frac{-4(1 + 512)}{3} = \frac{-4 \times 513}{3} = -684 \]

6. **Relate the Sum to \( \lambda \):**
We know from the problem statement that:
\[ S_9 = 4\lambda \implies -684 = 4\lambda \implies \lambda = \frac{-684}{4} = -171 \]

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