# The second and the fifth terms of a geometric sequence are 15 and 405, respectively. Is 32,805 a term of this sequence?

Nov 15, 2015

Yes

#### Explanation:

${A}_{n} = {A}_{1} {r}^{n - 1}$

${A}_{2} = {A}_{1} {r}^{2 - 1}$

$\implies {A}_{2} = {A}_{1} r = 15$

${A}_{5} = {A}_{1} {r}^{5 - 1}$

$\implies {A}_{5} = {A}_{1} {r}^{4} = 405$

${A}_{1} r {r}^{3} = 405$

$\implies 15 {r}^{3} = 405$

$\implies {r}^{3} = 27$

$\implies r = 3$

${A}_{2} = 15 = {A}_{1} r$
$\implies {A}_{1} \cdot 3 = 15$
$\implies {A}_{1} = 5$

For 32805 to be a term of the sequence, it must satisfy the equation

${A}_{n} = 5 \cdot {3}^{n - 1}$

where $n \in \mathbb{N}$

$32805 = 5 \cdot {3}^{n - 1}$

$\implies 6561 = {3}^{n - 1}$

$\implies {3}^{8} = {3}^{n - 1}$

$\implies 8 = n - 1$

$\implies n = 9$

Since $9 \in \mathbb{N}$, 32805 is a term of the sequence.