Step 1: Understanding the Question:
The problem involves an infinite geometric progression (GP). We are given two specific pieces of information: the sum of all terms in the series up to infinity and the value of the second term in that series. Our goal is to solve for the two defining parameters of any geometric sequence: the starting term ($a$) and the constant multiplier or common ratio ($r$). For the sum of an infinite series to exist, we must assume that the absolute value of the ratio is less than one ($|r|<1$).
Step 2: Key Formulas and approach:
Two fundamental formulas for geometric sequences are required:
1. The sum of an infinite GP: $S_{\infty} = \frac{a}{1 - r}$.
2. The $n$-th term of a GP: $T_n = a \cdot r^{(n-1)}$. Therefore, the second term ($T_2$) is $a \cdot r$.
The approach involves setting up a system of two equations based on the given values ($S_{\infty} = 4$ and $T_2 = 0.75$) and solving them simultaneously, typically by expressing one variable in terms of the other.
Step 3: Detailed Explanation:
We start by writing the equations based on the problem statement:
Equation (1) from the sum: $\frac{a}{1 - r} = 4 \implies a = 4(1 - r)$.
Equation (2) from the second term: $ar = \frac{3}{4}$.
We substitute the expression for $a$ from Eq (1) into Eq (2):
$[4(1 - r)] \cdot r = \frac{3}{4}$.
Distribute the $r$: $4r - 4r^2 = \frac{3}{4}$.
Multiply the entire equation by 4 to clear the fraction: $16r - 16r^2 = 3$.
Rearrange into a standard quadratic form: $16r^2 - 16r + 3 = 0$.
We solve this by factoring: $16r^2 - 12r - 4r + 3 = 0$.
Grouping terms: $4r(4r - 3) - 1(4r - 3) = 0 \implies (4r - 1)(4r - 3) = 0$.
This gives $r = 1/4$ or $r = 3/4$.
Case 1: If $r = 1/4$, then $a = 4(1 - 1/4) = 4(3/4) = 3$.
Case 2: If $r = 3/4$, then $a = 4(1 - 3/4) = 4(1/4) = 1$.
Comparing with the options, $(a=3, r=1/4)$ is present as option (D).
Step 4: Final Answer:
The parameters that satisfy both the sum and the second term conditions are $a = 3$ and $r = 1/4$. Thus, the correct option is (D).