Temos uma progressão geométrica (PG) decrescente onde:
[tex]a_1=16\\ a_5=9[/tex]
Podemos calcular a razão através do termo geral de uma PG:
[tex]\boxed{a_n=a_kq^{n-k}}[/tex]
[tex]a_5=a_1q^4\ \therefore\ \boxed{q=\sqrt[4]{\dfrac{a_5}{a_1}}}=\sqrt[4]{\dfrac{3^2}{2^4}}\ \therefore\ \boxed{q=\pm\dfrac{\sqrt{3}}{2}}[/tex]
Como a PG é decrescente (e aparentemente não-oscilante):
[tex]q>0\ \therefore\ \boxed{q=\dfrac{\sqrt{3}}{2}}\ \to\ \boxed{q=\dfrac{3^{\frac{1}{2}}}{2}}[/tex]
O termo geral dessa PG será:
[tex]a_n=a_1q^{n-1}=16\bigg(\dfrac{3^{\frac{1}{2}}}{2}\bigg)^{n-1}=\dfrac{2^4}{2^{n-1}}\bigg(3^{\frac{n-1}{2}}\bigg)=[/tex]
[tex]\dfrac{1}{2^{n-1-4}}3^{\frac{n-1}{2}}\ \therefore\ \boxed{a_n=\dfrac{3^{\frac{n-1}{2}}}{2^{n-5}}}[/tex]
Portanto, o septuagésimo termo será igual a:
[tex]a_{70}=\dfrac{3^{\frac{70-1}{2}}}{2^{70-5}}=\dfrac{3^{\frac{68}{2}+\frac{1}{2}}}{2^{65}}\ \therefore\ \boxed{a_{70}=\dfrac{3^{34}\sqrt{3}}{2^{65}}}[/tex]
