JEE Mains · Maths · STD 11 - 8. sequence and series
If \(a, b\) and \(c\) be three distinct numbers in \(G.P.\) and \(a + b + c = xb\) then \(x\) can not be
- A \(-2\)
- B \(-3\)
- C \(4\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\(a + ar + a{r^2} = xar\) since \(a \ne 0\,\,\) so \(\frac{{{r^2}r + 1}}{r} = x\); \(1 + r + \frac{1}{r} = x\) \(r + \frac{1}{r} \in ( - \infty , - 2] \cup [2,\infty )\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x \in ( - \infty , - 1] \cup [3,\infty )\)
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