The value of $(0.16)^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots . to \infty\right)}$ is equal to

  • [JEE MAIN 2020]
  • A

    $-4$

  • B

    $2$

  • C

    $-2$

  • D

    $4$

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