# How do you write a rule for the nth term of the geometric term given the two terms a_4=-8/9, a_7=-64/243?

Feb 28, 2017

${a}_{n} = - {2}^{n - 1} / {3}^{n - 2}$

#### Explanation:

Suppose using standard notation for a GP sequence that the first term is $a$ and the common ratio is $r$, the first few terms of the sequence are:

$\left(a , a r , a {r}^{2} , a {r}^{3} , a {r}^{4} , \ldots\right\}$

Assuming that the first term is ${a}_{1}$, then;

${a}_{1} = a$
${a}_{2} = a r$
${a}_{3} = a {r}^{2}$
$\vdots$
${a}_{n} = a {r}^{n - 1}$

Then ${a}_{4} = - \frac{8}{9} \setminus \setminus \setminus \implies a {r}^{3} = - \frac{8}{9} \setminus \setminus \setminus \setminus \setminus \setminus \ldots . . \left[1\right]$
And, ${a}_{7} = - \frac{64}{243} \implies a {r}^{6} = - \frac{64}{243} \setminus \setminus \setminus \ldots . . \left[2\right]$

$E q \left[2\right] \div i \mathrm{de} E q \left[1\right]$ gives;

$\setminus \setminus \frac{a {r}^{6}}{a {r}^{3}} = \frac{- \frac{64}{243}}{- \frac{8}{9}}$
$\therefore {r}^{3} = \frac{64}{243} \cdot \frac{9}{8}$
$\therefore {r}^{3} = \frac{8}{27}$
$\therefore \setminus \setminus r = \frac{2}{3}$

Subs ${r}^{3} = \frac{8}{27}$ into $E q \left[1\right]$ gives:

$a \cdot \frac{8}{27} = - \frac{8}{9}$
$\therefore a = - \frac{8}{9} \cdot \frac{27}{8}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 3$

And so the terms form a GP with $a = - 3$, $r = \frac{2}{3}$

So the ${n}^{t h}$ term is given by:

${a}_{n} = a {r}^{n - 1}$
$\setminus \setminus \setminus \setminus = - 3 {\left(\frac{2}{3}\right)}^{n - 1}$
$\setminus \setminus \setminus \setminus = - {2}^{n - 1} / {3}^{n - 2}$

Check

$n = 4 \implies {a}_{4} = - {2}^{3} / {3}^{2} = - \frac{8}{9}$
$n = 7 \implies {a}_{7} = - {2}^{6} / {3}^{5} = - \frac{64}{243}$