Question

If 9, 6, p are in arithmetic progression, 9, 6, q are in geometric progression and 9, 6, are in harmonic progression, then what is the value of (4p – 6q + r)?

• A. 3
• B. 4.5
• C. 12
• D. 16
• E. 15.5

Answer 1

The Correct Option is B

To find the value of \(4p - 6q + r\) given that 9, 6, p are in arithmetic progression (AP), 9, 6, q are in geometric progression (GP), and 9, 6, r are in harmonic progression (HP), we need to understand the properties of AP, GP, and HP and solve for p, q, and r accordingly. 

1. In an Arithmetic Progression (AP), the difference between consecutive terms is constant. So, for 9, 6, p: \(d = 6 - 9 = -3\) (common difference) Using the formula for a three-term AP, we have: \(6 - 9 = p - 6 \Rightarrow -3 = p - 6 \Rightarrow p = 3\).
2. In a Geometric Progression (GP), the ratio of consecutive terms is constant. So, for 9, 6, q: \(r = \frac{6}{9} = \frac{2}{3}\) (common ratio) Using the formula for a three-term GP, we have: \(\frac{6}{9} = \frac{q}{6} \Rightarrow q = \frac{6 \times 2}{3} = 4\).
3. In a Harmonic Progression (HP)\frac{1}{9}, \(\frac{1}{6}\), \(\frac{1}{r}\). These should be in arithmetic progression. Therefore: \(\frac{1}{6} - \frac{1}{9} = \frac{1}{r} - \frac{1}{6}\). Solving for \(\frac{1}{r}\), we have: \[ \begin{align*} \frac{1}{6} - \frac{1}{9} &= \frac{1}{r} - \frac{1}{6} \\ \frac{3}{54} &= \frac{1}{r} - \frac{1}{6} \\ \frac{1}{18} &= \frac{1}{r} - \frac{1}{6} \\ \frac{1}{r} &= \frac{1}{18} + \frac{1}{6} = \frac{1 + 3}{18} = \frac{4}{18} = \frac{2}{9} \\ r &= \frac{9}{2} = 4.5\ . \end{align*} \]

Therefore, the value of \(4p - 6q + r\) is -7.5, not among the provided options.