Find the 6th and the nth terms of the list of numbers $\dfrac{3}{2}, \dfrac{3}{4}, \dfrac{3}{8}, …$

The given list of numbers

$\dfrac{3}{2}, \dfrac{3}{4}, \dfrac{3}{8}, …$ is a G.P. with first term a = $\dfrac{3}{2}$ and common ratio = r = $\dfrac{1}{2}.$

By formula, a_n = ar^{n - 1}

$a$6 = \dfrac{3}{2}\Big(\dfrac{1}{2}\Big)^{6 - 1} = \dfrac{3}{2}\Big(\dfrac{1}{2}\Big)^5 = \dfrac{3}{2} \times \dfrac{1}{32} = \dfrac{3}{64}. \\[1em] an = \dfrac{3}{2}\Big(\dfrac{1}{2}\Big)^{n - 1} = \dfrac{3}{2 \times 2^n \times 2^{-1}} = \dfrac{3}{2^n}.

Hence, the 6th term is $\dfrac{3}{64} \text{ and nth term is } \dfrac{3}{2^n}.$