Key Concepts and Formulas

• A Geometric Progression (GP) is a sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio ($r$). The $n$-th term is given by $T_n = ar^{n-1}$, where $a$ is the first term.
• The condition "alternately positive and negative" for a GP implies that the common ratio $r$ must be negative.

Step-by-Step Solution

Step 1: Define the terms of the GP and set up equations from the problem statement. Let the first term of the geometric progression be $a$ and the common ratio be $r$. The terms are $T_1 = a$, $T_2 = ar$, $T_3 = ar^2$, and $T_4 = ar^3$. We are given:

1. The sum of the first two terms is 12: $T_1 + T_2 = 12$ $a + ar = 12$ Factoring out $a$, we get: $a(1 + r) = 12 \quad \text{(Equation 1)}$
2. The sum of the third and fourth terms is 48: $T_3 + T_4 = 48$ $ar^2 + ar^3 = 48$ Factoring out $ar^2$, we get: $ar^2(1 + r) = 48 \quad \text{(Equation 2)}$

Step 2: Solve for the common ratio ($r$) by dividing Equation 2 by Equation 1. Dividing Equation 2 by Equation 1 allows us to eliminate $a$ and $(1+r)$: $\frac{ar^2(1 + r)}{a(1 + r)} = \frac{48}{12}$ Assuming $a \neq 0$ and $1+r \neq 0$ (which are necessary for the sums to be non-zero), we can cancel the common terms: $r^2 = 4$ Taking the square root of both sides, we get: $r = \pm 2$

Step 3: Use the condition of alternating signs to determine the correct value of $r$. The problem states that the terms of the geometric progression are alternately positive and negative. This occurs if and only if the common ratio $r$ is negative.

• If $r > 0$, all terms would have the same sign as the first term $a$.
• If $r < 0$, the signs of the terms will alternate (e.g., $a, ar, ar^2, ar^3, ...$ would be $+, -, +, -, ...$ or $-, +, -, +, ...$). Therefore, we must choose $r = -2$.

Step 4: Substitute the value of $r$ back into Equation 1 to solve for the first term ($a$). Using Equation 1: $a(1 + r) = 12$ Substitute $r = -2$: $a(1 + (-2)) = 12$ $a(-1) = 12$ $a = -12$

Step 5: Verify the solution with the given conditions. With $a = -12$ and $r = -2$, the GP is: $T_1 = -12$ $T_2 = (-12)(-2) = 24$ $T_3 = (-12)(-2)^2 = -48$ $T_4 = (-12)(-2)^3 = 96$ The sequence is $-12, 24, -48, 96, ...$, which has alternating signs. Sum of the first two terms: $T_1 + T_2 = -12 + 24 = 12$ (Correct). Sum of the third and fourth terms: $T_3 + T_4 = -48 + 96 = 48$ (Correct). All conditions are satisfied.

Common Mistakes & Tips

• Forgetting the alternating sign condition: This condition is crucial for selecting the correct value of $r$ from $r^2=4$. If ignored, one might incorrectly proceed with $r=2$.
• Algebraic errors: Be careful when factoring and dividing equations, especially with negative signs.
• Verification: Always check your final values of $a$ and $r$ against all conditions given in the problem to ensure accuracy.

Summary The problem involves a geometric progression where the sums of consecutive pairs of terms are given. By setting up equations based on the definition of a GP and the given sums, we can derive a quadratic equation for the common ratio $r$. The condition of alternating signs is essential for selecting the correct negative value of $r$. Substituting this value back into one of the initial equations allows us to find the first term $a$.

The final answer is $\boxed{-12}$.