How do you write a rule for the nth term of the geometric sequence given the two terms a_2=-153, a_4=-17?

1 Answer
Jan 19, 2018

${u}_{n} = \pm 459 \cdot {\left(\pm \frac{1}{3}\right)}^{n - 1}$

Explanation:

geometric sequence: ${u}_{n} = a {r}^{-} 1$

where $a$ is the starting term

and $r$ is the number by which one number is multiplied to make the next number in the sequence. (common ratio)

${u}_{2} = - 153$
${u}_{4} = - 17$

${u}_{2} : n = 2$

${u}_{2} = a {r}^{2 - 1} = a {r}^{1}$

${u}_{2} = a r$

${u}_{4} : n = 4$

${u}_{4} = a {r}^{4 - 1} = a {r}^{3}$

$a r = - 153$
$a {r}^{3} = - 17$

${r}^{2} = \frac{a {r}^{3}}{a r} = \frac{- 17}{-} 153$

${r}^{2} = \frac{1}{9}$

$r = \pm \frac{1}{3}$

$a r = - 153$

$r = \pm \frac{1}{3}$

$a = - \frac{153}{\pm \frac{1}{3}} = - 153 \cdot \pm 3$

$a = \pm 459$

the $n$th term of the geometric sequence is ${u}_{n} = \pm 459 \cdot {\left(\pm \frac{1}{3}\right)}^{n - 1}$