Question

O primeiro termo de uma P.G é 5√2, a razão √2 e o último termo é 80. Calcule:

a) quantos termos têm essa P.G;
b) O seu quinto termo;​

Answer 1

a)

Utilizando a equação do termo geral da PG, podemos determinar o numero de termos dessa PG:

$a_n~=~a_1~.~q^{n-1}\\\\\\80~=~5\sqrt{2}~.~\sqrt{2}^{~n-1}\\\\\\\dfrac{80}{5\sqrt{2}}~=~\sqrt{2}^{~n-1}\\\\\\\dfrac{16}{\sqrt{2}}~=~\sqrt{2}^{~n-1}\\\\\\\dfrac{16}{\sqrt{2}}~.~\dfrac{\sqrt{2}}{\sqrt{2}}~=~\sqrt{2}^{~n-1}\\\\\\\dfrac{16\sqrt{2}}{2}~=~\sqrt{2}^{~n-1}\\\\\\8\sqrt{2}~=~\sqrt{2}^{~n-1}\\\\\\2^3.2^{\frac{1}{2}}~=~\left(2^{\frac{1}{2}}\right)^{n-1}\\\\\\2\!\!\!\backslash^{3+\frac{1}{2}}~=~2\!\!\!\backslash^{\frac{1}{2}.(n-1)}\\\\\\3+\frac{1}{2}~=~\frac{1}{2}.(n-1)$

$\frac{7}{2}~=~\frac{1}{2}.(n-1)\\\\\\n-1~=~\frac{7}{2}.\frac{2}{1}\\\\\\n~=~7+1\\\\\\\boxed{n~=~8~termos}$

b)

Utilizando novamente a equação do termo geral, podemos determinar o valor do 5° termo:

$a_5~=~a_1~.~q^{5-1}\\\\\\a_5~=~5\sqrt{2}~.~\sqrt{2}^{~4}\\\\\\a_5~=~5\sqrt{2}~.~2^2\\\\\\a_5~=~5\sqrt{2}~.~4\\\\\\\boxed{a_5~=~20\sqrt{2}}$