Question

If \(a, b, c\) be three distinct real numbers in G.P. and \(a + b + c = xb\), then \(x\) cannot be

• A. \(2\)
• B. \(-3\)
• C. \(4\)
• D. \(-2\)

Hint:

Write the terms as \(b/r, b, br\) and express \(x\) in terms of \(r\); then find the range of \(r + 1/r\) for real \(r\).

Answer 1

The Correct Option is A

Let the G.P. terms be \(a = \frac{b}{r},\ b,\ c = br\) with \(r \ne 0, 1\) (real and distinct, \(b \ne 0\)).
\(a + b + c = b\left(\frac{1}{r} + 1 + r\right) = xb \Rightarrow x = 1 + r + \frac{1}{r}\).
For real \(r \gt 0\): \(r + \frac{1}{r} \ge 2\), so \(x \ge 3\). For real \(r \lt 0\): \(r + \frac{1}{r} \le -2\), so \(x \le -1\).
Hence \(x\) can never lie in the interval \((-1, 3)\). Among the options only \(x = 2\) lies in it (it would need \(r^2 - r + 1 = 0\), which has no real root).
\(-3, 4, -2\) are all attainable, so \(x\) cannot be \(2\).
Answer: (1) \(2\)