Find the G.P. whose 4th and 7th terms are $\dfrac{1}{18}$ and $-\dfrac{1}{486}$ respectively.

Let a be the first term and r be the common ratio.

We know that,

nth term of a G.P. is given by,

T_n = ar^{n - 1}

Given,

⇒ 4th term of G.P is $\dfrac{1}{18}$

⇒ ar^{4 - 1} = $\dfrac{1}{18}$

⇒ ar³ = $\dfrac{1}{18}$ ….(1)

Given,

⇒ 7th term of G.P. is $-\dfrac{1}{486}$

⇒ ar^{7 - 1} = $-\dfrac{1}{486}$

⇒ ar⁶ = $-\dfrac{1}{486}$….(2)

Divide Equation 2 by Equation 1:

$\Rightarrow \dfrac{ar^6}{ar^3} = \dfrac{-\dfrac{1}{486}}{\dfrac{1}{18}} \\[1em] \Rightarrow r^{6 - 3} = -\dfrac{1}{486} \times 18 \\[1em] \Rightarrow r^{3} = -\dfrac{1}{486} \times 18 \\[1em] \Rightarrow r^{3} = -\dfrac{1}{27} \\[1em] \Rightarrow r = \sqrt[3]{-\dfrac{1}{27}} \\[1em] \Rightarrow r = -\dfrac{1}{3}.$

Substituting $r = -\dfrac{1}{3}$ into Equation 1, we get:

$\Rightarrow a\Big(-\dfrac{1}{3}\Big)^3 = \dfrac{1}{18} \\[1em] \Rightarrow a\Big(-\dfrac{1}{27}\Big) = \dfrac{1}{18} \\[1em] \Rightarrow a = -27 \times \dfrac{1}{18} \\[1em] \Rightarrow a = -\dfrac{3}{2}.$

G.P. is,

$\Rightarrow -\dfrac{3}{2}, -\dfrac{3}{2} \times \Big(-\dfrac{1}{3}\Big), -\dfrac{3}{2} \times \Big(-\dfrac{1}{3}\Big)^2, …… \\[1em] \Rightarrow -\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{1}{6}, ……..$

Hence, the G.P. is $-\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{1}{6}, ……..$