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Step-by-Step Solution
Step 1: Express the given conditions in terms of $a_1$ and $r$
We are given a geometric progression (GP) $a_1, a_2, a_3, \dots$ with common ratio $r$. The first two conditions can be written as:
$a_1 + a_2 = 4$
$a_3 + a_4 = 16$
Since $a_2 = a_1 \cdot r$ and $a_3 = a_1 \cdot r^2$, $a_4 = a_1 \cdot r^3$, we get:
(1) $a_1 + a_1 r = 4$, which implies $a_1(1 + r) = 4$.
(2) $a_1 r^2 + a_1 r^3 = 16$, which implies $a_1 r^2 (1 + r) = 16$.
Step 2: Determine the common ratio $r$
Divide the first condition by the second:
$$
\frac{a_1 (1 + r)}{a_1 r^2 (1 + r)}
= \frac{4}{16}
\quad \Longrightarrow \quad
\frac{1}{r^2} = \frac{1}{4}
\quad \Longrightarrow \quad
r^2 = 4
\quad \Longrightarrow \quad
r = \pm 2.
$$
Step 3: Identify $a_1$ given $a_1 < 0$
Since $a_1 < 0$, we choose $r = -2$. Then from $a_1(1 + r) = 4$, substitute $r = -2$:
$$
a_1 (1 - 2) = 4
\quad \Longrightarrow \quad
a_1 \cdot (-1) = 4
\quad \Longrightarrow \quad
a_1 = -4.
$$
Step 4: Sum of first 9 terms of the GP
The sum of first $n$ terms of a GP with first term $a_1$ and common ratio $r$ is given by:
$$
S_n = a_1 \cdot \frac{r^n - 1}{r - 1}.
$$
For $n = 9$, $a_1 = -4$, and $r = -2$, we have:
$$
S_9 = -4 \cdot \frac{(-2)^9 - 1}{-2 - 1}
= -4 \cdot \frac{-512 - 1}{-3}
= -4 \cdot \frac{-513}{-3}.
$$
Carefully evaluating inside:
$(-2)^9 = -512$, so $(-2)^9 - 1 = -512 - 1 = -513$.
Hence,
$$
S_9
= -4 \cdot \frac{-513}{-3}
= -4 \cdot \frac{-513}{-3}.
$$
Observe that $\frac{-513}{-3} = 171,
$$
\text{so } S_9 = -4 \times 171 = -684.
$$
Step 5: Relate $S_9$ to $4\lambda$ and find $\lambda$
It is given that $S_9 = 4\lambda$. We have found $S_9 = -684$, so:
$$
-684 = 4\lambda
\quad \Longrightarrow \quad
\lambda = \frac{-684}{4} = -171.
$$
Final Answer
Hence, the value of $\lambda$ is $\boxed{-171}$.
