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Let \(t_1, t_2, t_3, \ldots, t_{2n-2}, t_{2n-1}, t_{2n}\) be in G.P. with common ratio \(r\) . Then

  • A.\(t_1, t_3, t_5, \ldots, t_{2n-5}, t_{2n-3}, t_{2n-1}\) are in G.P. with common ratio \(r\)
  • B.\(t_1, t_4, t_7, \ldots, t_{2n-7}, t_{2n-4}, t_{2n-1}\) are in G.P. with common ratio \(r^{2}\)
  • C.\(t_1, t_3, t_5, \ldots, t_{2n-5}, t_{2n-3}, t_{2n-1}\) are in G.P. with common ratio \(r^{2}\)
  • D.\(t_2, t_4, t_6, \ldots, t_{2n-4}, t_{2n-2}, t_{2n}\) are in G.P. with common ratio \(r^{3}\)
  • E.\(t_2, t_4, t_6, \ldots, t_{2n-4}, t_{2n-2}, t_{2n}\) are in G.P. with common ratio \(r^{5}\)

Answer: C

The odd-numbered terms \(t_1, t_3, t_5, \ldots, t_{2n-1}\) are in G.P. with common ratio \(r^{2}\).

Each term is \(t_k = t_1 r^{k-1}\). Moving from one odd-numbered term to the next skips two places:

\[ \begin{aligned} \frac{t_{k+2}}{t_k} &= \frac{t_1 r^{k+1}}{t_1 r^{k-1}} \\ &= r^{2} \end{aligned} \]

So the ratio is \(r^{2}\), not \(r\), which rules out the first option. Terms \(t_1, t_4, t_7, \ldots\) skip three places and have ratio \(r^{3}\), not \(r^{2}\). The even-numbered terms also have ratio \(r^{2}\), not \(r^{3}\) or \(r^{5}\).

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