Question

ache o 9 termo da pg (40,20,10...)

Answer 1

$a_{1}=40\\a_{2}=20\\\\q=\dfrac{a_{2}}{a_{1}}=\dfrac{20}{40}=\dfrac{1}{2}$
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$a_{n}=a_{1}\cdot q^{n-1}\\a_{9}=a_{1}\cdot q^{9-1}\\a_{9}=a_{1}\cdot q^{8}$

Substituindo os valores:

$a_{9}=40\cdot\left(\frac{1}{2}\right)^{8}\\\\a_{9}=40\cdot\frac{1}{256}\\\\a_{9}=5\cdot\frac{1}{32}\\\\\boxed{\boxed{a_{9}=\frac{5}{32}}}$

Answer 2

Utilize a Fórmula do termo geral de uma PG:

$a_n=a_1.q^{n-1}\\ \\ a_9=40.(\frac{1}{2})^8=40.\frac{1}{256}=\frac{5}{32}$