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Step-by-Step Solution
Step 1: Express the first two terms and their sum
Let the first term of the geometric progression (GP) be $a$, and the common ratio be $r$.
Then the first two terms are:
• First term: $a$
• Second term: $ar$
According to the problem, these two terms add to 12:
$$
a + ar = 12 \quad \Longrightarrow \quad a(1 + r) = 12 \,.\!
$$
Step 2: Express the third and fourth terms and their sum
The third and fourth terms are:
• Third term: $ar^2$
• Fourth term: $ar^3$
Their sum is given as 48:
$$
ar^2 + ar^3 = 48 \quad \Longrightarrow \quad a\,r^2\bigl(1 + r\bigr) = 48 \,.
$$
Step 3: Divide the second equation by the first to find $r^2$
Divide
$$
a\,r^2\,(1 + r) = 48
$$
by
$$
a\,(1 + r) = 12\,,
$$
assuming $1 + r \neq 0$ and $a \neq 0$:
$$
\frac{a\,r^2\,(1 + r)}{a\,(1 + r)} = \frac{48}{12} \quad\Longrightarrow\quad r^2 = 4 \,.
$$
Hence,
$$
r = \pm 2\,.
$$
Step 4: Use the alternating-sign condition to determine $r$
The problem states that the terms of the GP are alternately positive and negative.
For consecutive terms to flip sign each time, $r$ must be negative:
$$
r = -2\,.
$$
Step 5: Solve for the first term $a$
Substitute $r = -2$ into the sum of the first two terms:
$$
a(1 + r) = a\,\bigl(1 + (-2)\bigr) = a \times (-1) = 12\,.
$$
Hence,
$$
a = -12\,.
$$
Step 6: Conclusion
Although the provided “Correct Answer” in the question statement is listed as 12, the consistent solution (given the alternating signs and the two sums) mathematically yields
$$
a = -12.
$$
Thus, the first term that satisfies all the given conditions (sum of first two terms is 12, sum of third and fourth terms is 48, and alternating signs) is
$$
-12.
$$
