The 11th term of the G.P. $\dfrac{1}{8}, -\dfrac{1}{4}, 2, -1, ….$ is

1. 64

2. -64

3. 128

4. -128

Given, a = $\dfrac{1}{8}$ and common ratio = r = $\dfrac{-\dfrac{1}{4}}{\dfrac{1}{8}} = -2.$

We know that

    a_n = ar^{n - 1}

$\therefore a_{11} = \dfrac{1}{8}(-2)^{11 - 1} \\[1em] = \dfrac{1}{8}(-2)^{10} \\[1em] = \dfrac{1}{8} \times 2^{10} \\[1em] = \dfrac{2^{10}}{2^3} \\[1em] = 2^7 \\[1em] = 128.$

Hence, Option 3 is the correct option.