Question

Dada um P.G ( 4, -6, ...) determine o sexto termo desse progressão

Answer 1

A seguinte progressão geométrica é dada

$(4,\,-6,\,\ldots)$


Encontrando a razão da P.G.:

$q=\dfrac{a_2}{a_1}\\\\\\ q=\dfrac{-6}{4}\\\\\\ \boxed{\begin{array}{c} q=-\,\dfrac{3}{2} \end{array}}$


O termo geral desta P.G. é dado por

$a_n=a_1\cdot q^{n-1}\\\\ \boxed{\begin{array}{c} a_n=4\cdot \left(-\,\dfrac{3}{2} \right )^{\!\!n-1} \end{array}}~~~~~~\text{com }n=1,\,2,\,3,\,\ldots$


O sexto termo é encontrado fazendo $n=6:$

$a_6=4\cdot \left(-\,\dfrac{3}{2} \right )^{\!\!6-1}\\\\\\ a_6=4\cdot \left(-\,\dfrac{3}{2} \right )^{\!\!5}\\\\\\ a_6=4\cdot \dfrac{(-1)^5\cdot 3^5}{2^5}\\\\\\ a_6=4\cdot \dfrac{(-1)^5\cdot 3^5}{2^2\cdot 2^3}\\\\\\ a_6=\diagup\!\!\!\! 4\cdot \dfrac{(-1)^5\cdot 3^5}{\diagup\!\!\!\! 4\cdot 2^3}$

$a_6=\dfrac{(-1)^5\cdot 3^5}{2^3}\\\\\\ a_6=\dfrac{(-1)\cdot 243}{8}\\\\\\ \boxed{\begin{array}{c} a_6=-\,\dfrac{243}{8} \end{array}}$