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:
We are dealing with an infinite geometric progression. We are given its sum to infinity and the value of its second term. We need to find the first term (\(a\)) and the common ratio (\(r\)).


Step 2: Key Formula or Approach:
The sum to infinity of a geometric series is \(S_\infty = \frac{a}{1 - r}\) (where \(|r| < 1\)).
The \(n\)-th term of a geometric series is \(T_n = a r^{n-1}\). Thus, the second term is \(T_2 = ar\).


Step 3: Detailed Explanation:
From the given information, we can set up two equations: 1) \(S_\infty = \frac{a}{1 - r} = 4 \implies a = 4(1 - r)\) 2) \(T_2 = ar = \frac{3}{4} \implies a = \frac{3}{4r}\)
Equating the two expressions for \(a\): \[ 4(1 - r) = \frac{3}{4r} \] Multiply both sides by \(4r\): \[ 16r(1 - r) = 3 \] \[ 16r - 16r^2 = 3 \] Rearrange into a standard quadratic equation: \[ 16r^2 - 16r + 3 = 0 \] Factorizing the quadratic equation: \[ 16r^2 - 12r - 4r + 3 = 0 \] \[ 4r(4r - 3) - 1(4r - 3) = 0 \] \[ (4r - 1)(4r - 3) = 0 \] This gives two possible values for \(r\): \(r = \frac{1}{4}\) or \(r = \frac{3}{4}\)
Now, let's find the corresponding values for \(a\) using \(a = \frac{3}{4r}\): If \(r = \frac{1}{4}\): \[ a = \frac{3}{4(\frac{1}{4})} = \frac{3}{1} = 3 \] This corresponds to pair \((a = 3, r = \frac{1}{4})\), which matches option D.
If \(r = \frac{3}{4}\): \[ a = \frac{3}{4(\frac{3}{4})} = \frac{3}{3} = 1 \] This gives pair \((a = 1, r = \frac{3}{4})\), which is not among the options.


Step 4: Final Answer:
The correct values are \(a = 3, r = \frac{1}{4}\).

Answer 2

Since this question gives four candidate pairs of \((a,r)\), one efficient alternate approach is to test each pair directly against the two given conditions, the sum to infinity \(S_\infty=\frac{a}{1-r}=4\) and the second term \(T_2=ar=\frac{3}{4}\), rather than solving the system algebraically from scratch.

1. \(a=\frac{7}{4}, r=\frac{3}{7}\): Sum check: \(\frac{7/4}{1-3/7}=\frac{7/4}{4/7}=\frac{49}{16}\neq4\). This pair fails the sum condition.
2. \(a=2, r=\frac{3}{8}\): Sum check: \(\frac{2}{1-3/8}=\frac{2}{5/8}=\frac{16}{5}=3.2\neq4\). This pair fails the sum condition.
3. \(a=\frac{3}{2}, r=\frac{1}{2}\): Sum check: \(\frac{3/2}{1-1/2}=\frac{3/2}{1/2}=3\neq4\). This pair fails the sum condition.
4. \(a=3, r=\frac{1}{4}\): Sum check: \(\frac{3}{1-1/4}=\frac{3}{3/4}=4\), which matches. Second term check: \(ar=3\times\frac{1}{4}=\frac{3}{4}\), which also matches.

Only the fourth pair satisfies both the sum-to-infinity condition and the second-term condition simultaneously; every other pair fails at least the sum check.

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