Question

Consider an infinite series with first term a and common ratio r. If its sum is 4 and the second term is \(\frac{3}{4}\), then

• A. \(a = \frac{7}{4}, r = \frac{3}{7}\)
• B. \(a = 2, r = \frac{3}{8}\)
• C. \(a = \frac{3}{2}, r = \frac{1}{2}\)
• D. \(a = 3, r = \frac{1}{4}\)

Hint:

When dealing with simultaneous equations from sequences, substituting \(a\) in terms of \(r\) usually leads straight to a solvable quadratic equation. Always check both roots against the given options.

Answer 1

The Correct Option is D

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).

Answer 2

Instead of eliminating \(a\) first as is usually done, we can flip the substitution and eliminate \(r\) instead, ending up with a quadratic in \(a\) rather than in \(r\).

1. Setting up from the second term: From \(T_2=ar=\frac{3}{4}\), we can express \(r=\frac{3}{4a}\).
2. Substituting into the sum condition: The sum condition is \(\frac{a}{1-r}=4\), which rearranges to \(a=4(1-r)=4-4r\). Substituting \(r=\frac{3}{4a}\) gives \(a=4-4\left(\frac{3}{4a}\right)=4-\frac{3}{a}\).
3. Forming a quadratic in \(a\): Multiply through by \(a\): \(a^2=4a-3\), which rearranges to \(a^2-4a+3=0\). Factoring gives \((a-1)(a-3)=0\), so \(a=1\) or \(a=3\).
4. Finding the matching \(r\) for each \(a\): If \(a=3\), then \(r=\frac{3}{4(3)}=\frac{1}{4}\), giving the pair \((3,\frac{1}{4})\), which appears among the options. If \(a=1\), then \(r=\frac{3}{4(1)}=\frac{3}{4}\), giving the pair \((1,\frac{3}{4})\), which does not appear among the options.

Only the pair from \(a=3\) matches one of the listed options.

Therefore, the correct answer is \(a=3, r=\frac{1}{4}\).