# Question #730a0

Mar 30, 2017

$27$

#### Explanation:

Denote the $n$:th element in the sequence ${a}_{n} = b {x}^{n - 1}$.
Then ${a}_{3} = b {x}^{2}$ and ${a}_{6} = b {x}^{5}$. Knowing that ${a}_{3} = 3$ and ${a}_{6} = \frac{1}{9}$, we get that
$3 = b \cdot {x}^{2}$
$\frac{1}{9} = b \cdot {x}^{5} = \left(b {x}^{2}\right) {x}^{3}$
Substituting the first line into the second we get that
$\frac{1}{9} = 3 {x}^{3}$, which gives that
$x = \frac{1}{3}$.
Inserting into the first equation, we get that
$b = \frac{3}{{x}^{2}} = 27$.
Thus the first element ${a}_{1} = b {x}^{1 - 1} = b = 27$.

Mar 30, 2017

$27.$

#### Explanation:

Let the Geom. Seq. be $a , a r , a {r}^{2} , \ldots , a {r}^{n - 1} , \ldots$

By what is given, $a {r}^{2} = 3 , \mathmr{and} , a {r}^{5} = \frac{1}{9.}$

$\therefore \frac{a {r}^{5}}{a {r}^{2}} = \frac{1}{9} \div 3$.

$\therefore {r}^{3} = \frac{1}{27}$

$\therefore r = \frac{1}{3.}$

Now, $a {r}^{2} = 3 , r = \frac{1}{3} \Rightarrow a = 3 \div {r}^{2} = 3 \div {\left(\frac{1}{3}\right)}^{2} = 3 \div \frac{1}{9} = 27.$

$\therefore \text{ The First term is } 27.$