Q161 / 270 · NDA 2023 (Phase 2) · Mathematics

Let $t_1, t_2, t_3, \ldots$ be in GP. What is $\left(t_1 t_3 \ldots t_{21}\right)^{\frac{1}{11}}$ equal to?

  • A
    $t_{10}$
  • B
    $t_{10}^2$
  • Correct answer
    C
    $t_{11}$
  • D
    $t_{11}^2$

Solution

since, $t_1, t_2, t_3 \ldots \ldots$ in G.P Let $\quad t_n=a r^{n-1}$ Now, $\quad\left(t_1 . t_3 \ldots . . t_{21}\right)^{\frac{1}{11}}$ $\begin{aligned} & =\left(\mathrm{a}^{11} \mathrm{r}^{110}\right)^{\frac{1}{11}}=\mathrm{ar}^{10}=\mathrm{t}_{11} \end{aligned}$

Hence, the correct answer is option: 'C'.

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