How do you write the expression for the nth term of the geometric sequence a_1=16, a_4=27/4, n=3?

Feb 23, 2018

$n$ th term is ${a}_{1} \cdot {r}^{n - 1}$ and ${a}_{3} = 16$

Explanation:

In Geometric progression $n$ th term is ${a}_{1} \cdot {r}^{n - 1}$ where

${a}_{1}$ is first term and $r$ is the common ratio.

${a}_{1} = 16 , {a}_{4} = \frac{27}{4}$

$\therefore {a}_{4} = 16 \cdot {r}^{4 - 1} \mathmr{and} {a}_{4} = 16 {r}^{3} = \frac{27}{4} \mathmr{and} {r}^{3} = \frac{27}{64}$or

$r = \sqrt[3]{\frac{27}{64}} = \frac{3}{4}$. When $n = 3 \therefore {a}_{3} = 16 \cdot {\left(\frac{3}{4}\right)}^{3 - 1}$

or ${a}_{3} = 16 \cdot {\left(\frac{3}{4}\right)}^{3 - 1} = 16 \cdot {\left(\frac{3}{4}\right)}^{2} = 9$