Question

If \(k\), 6 and \(k+5\) are the first three terms of a geometric series, then the possible values of the common ratio are

• A. \(\frac{1}{2}, 2\)
• B. 2, -2
• C. \(\frac{1}{3}, 3\)
• D. \(\frac{3}{2}, \frac{-2}{3}\)
• E. \(\frac{2}{3}, \frac{-3}{2}\)

Hint:

Always check both roots of the quadratic equation. In G.P. problems, a negative value for \(k\) is perfectly valid and leads to an alternating series.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
If three numbers \(a, b, c\) are in G.P., then the square of the middle term equals the product of the other two (\(b^2 = ac\)). This allows us to solve for the unknown \(k\).

Step 2: Key Formula or Approach:
1. For G.P.: \(b^2 = ac\).
2. Common ratio \(r = \frac{b}{a}\).

Step 3: Detailed Explanation:
The terms are \(k, 6, k+5\). Since they are in G.P.:
\[ 6^2 = k(k+5) \]
\[ 36 = k^2 + 5k \]
\[ k^2 + 5k - 36 = 0 \]
Factoring the quadratic:
\[ k^2 + 9k - 4k - 36 = 0 \]
\[ k(k+9) - 4(k+9) = 0 \]
\[ (k+9)(k-4) = 0 \implies k = -9 \text{ or } k = 4 \]
Case 1: If \(k = 4\):
Terms are 4, 6, 9.
Ratio \(r = \frac{6}{4} = \frac{3}{2}\).
Case 2: If \(k = -9\):
Terms are -9, 6, -4.
Ratio \(r = \frac{6}{-9} = -\frac{2}{3}\).

Step 4: Final Answer:
The possible values of the common ratio are \(\frac{3}{2}\) and \(-\frac{2}{3}\).