The first two terms of a geometric progression add up to 12. The sum of the third and the fourth terms is 48. If the terms of the geometric progression are alternately positive and negative, then the first term is

Let the first term of the G.P. be $$a$$ and the common ratio be $$r$$.

First two terms: $$T_1 = a$$, $$T_2 = ar$$
Third and fourth terms: $$T_3 = ar^{2}$$, $$T_4 = ar^{3}$$

Given sums:
$$a + ar = 12 \quad -(1)$$
$$ar^{2} + ar^{3} = 48 \quad -(2)$$

Factor out common terms in each sum:

From $$(1):$$ $$a(1 + r) = 12$$

From $$(2):$$ $$ar^{2}(1 + r) = 48$$

Divide $$(2)$$ by $$(1)$$ to eliminate $$a(1 + r)$$:

$$\frac{ar^{2}(1 + r)}{a(1 + r)} = \frac{48}{12} \;\Longrightarrow\; r^{2} = 4$$
Hence $$r = \pm 2$$.

The terms are stated to be alternately positive and negative. Consecutive terms of a G.P. have opposite signs only when the common ratio is negative. Therefore

$$r = -2$$.

Substitute $$r = -2$$ into $$(1)$$:

$$a\bigl(1 + (-2)\bigr) = 12 \;\Longrightarrow\; a(-1) = 12 \;\Longrightarrow\; a = -12$$.

Thus the first term is $$-12$$.

Option B which is: $$-12$$