Leonel Schwartz

## Answered question

2022-10-03

Find the 7th term of the geometric sequence with the given terms ${a}_{4}=54,{a}_{5}=162$

### Answer & Explanation

Derick Ortiz

Beginner2022-10-04Added 11 answers

Since this is a geometric sequence we know the following
${a}_{n}={a}_{1}{r}^{n-1}$ noticed we started at 1 that why we subtract 1 from n
the ratio $r=\frac{{a}_{n}}{{a}_{n-1}}$
We are given
${a}_{4}=54\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}\text{}{a}_{5}=162$
We first, need to find r
$r=\frac{{a}_{5}}{{a}_{4}}=\frac{162}{54}=3$
Then we can use the formula ${a}_{n}={a}_{1}{r}^{n-1}$
(but instead of finding ${a}_{1}$, we will use ${a}_{4}$ we need to subtract 4 from the nth term
${a}_{7}={a}_{4}{r}^{7-4}$
${a}_{7}=\left(54\right){\left(3\right)}^{3}$
$\text{}{a}_{7}=1458$

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