Let $$a_1, a_2, a_3, \ldots$$ be a G.P. such that $$a_1 < 0$$, $$a_1 + a_2 = 4$$ and $$a_3 + a_4 = 16$$. If $$\sum_{i=1}^{9} a_i = 4\lambda$$, then $$\lambda$$ is equal to.

Let the first term of the geometric progression be $$a_1$$ and the common ratio be $$r$$. By definition of a G.P. we therefore have

$$a_2 = a_1 r, \quad a_3 = a_1 r^2, \quad a_4 = a_1 r^3,$$ and so on.

We are told that $$a_1<0$$ and that the following two relations hold:

$$a_1 + a_2 = 4, \qquad a_3 + a_4 = 16.$$

First we use the relation $$a_1 + a_2 = 4$$. Substituting $$a_2 = a_1 r$$ gives

$$a_1 + a_1 r = 4.$$

Factoring out $$a_1$$ we obtain

$$a_1(1 + r) = 4 \quad\Longrightarrow\quad a_1 = \frac{4}{1 + r}. \quad -(1)$$

Next we use the relation $$a_3 + a_4 = 16$$. Writing these terms in terms of $$a_1$$ and $$r$$ we have

$$a_1 r^2 + a_1 r^3 = 16.$$

Again factoring out $$a_1 r^2$$ gives

$$a_1 r^2(1 + r) = 16. \quad -(2)$$

Both equations (1) and (2) contain the common factor $$(1 + r)$$, so dividing equation (2) by equation (1) eliminates $$a_1(1 + r)$$ entirely. We get

$$\frac{a_1 r^2(1 + r)}{a_1(1 + r)} = \frac{16}{4} \quad\Longrightarrow\quad r^2 = 4.$$

Taking square roots gives two possibilities:

$$r = 2 \quad\text{or}\quad r = -2.$$

We still need to satisfy the condition $$a_1 < 0$$. From equation (1)

$$a_1 = \frac{4}{1 + r}.$$

If $$r = 2$$, then $$a_1 = \dfrac{4}{1 + 2} = \dfrac{4}{3} > 0,$$ violating $$a_1<0$$. Hence $$r = 2$$ is not acceptable.

Thus we must have $$r = -2$$. Substituting this value back into equation (1) gives

$$a_1 = \frac{4}{1 + (-2)} = \frac{4}{-1} = -4,$$

which indeed satisfies $$a_1 < 0$$.

Now we compute the sum of the first nine terms of the G.P. The formula for the sum of the first $$n$$ terms when the common ratio $$r \neq 1$$ is

$$S_n = a_1 \frac{1 - r^n}{1 - r}.$$

Taking $$n = 9$$, $$a_1 = -4$$ and $$r = -2$$ we have

$$S_9 = -4 \, \frac{1 - (-2)^9}{1 - (-2)}.$$

We evaluate each part carefully. First,

$${(-2)^9} = -2^9 = -512.$$

Hence

$$1 - (-2)^9 = 1 - (-512) = 1 + 512 = 513.$$

Next, the denominator is

$$1 - (-2) = 1 + 2 = 3.$$

Putting these values together we get

$$S_9 = -4 \cdot \frac{513}{3}.$$

Since $$\frac{513}{3} = 171,$$ the sum simplifies to

$$S_9 = -4 \times 171 = -684.$$

The question states that this same sum can be written as $$\sum_{i=1}^{9} a_i = 4\lambda,$$ so we set

$$4\lambda = -684.$$

Dividing both sides by 4 yields

$$\lambda = \frac{-684}{4} = -171.$$

Hence, the correct answer is Option B.