Question

em uma PG o 5° termo mais o 9° termo é igual a 40. o 7° mais o 11º é igual a 360. calcule a razão dessa PG

Answer 1

Vamos escrever todos termos em função de a1 e a razão (q) utilizando a equação do termo geral da PG:

$\boxed{a_n~=~a_1~.~q^{n-1}}\\\\\\\\\rightarrow~a_5~=~a_1~.~q^{5-1}~~~\rightarrow~~~\boxed{a_5~=~a_1~.~q^4}\\\\\rightarrow~a_9~=~a_1~.~q^{9-1}~~~\rightarrow~~~\boxed{a_9~=~a_1~.~q^8}\\\\\rightarrow~a_7~=~a_1~.~q^{7-1}~~~\rightarrow~~~\boxed{a_7~=~a_1~.~q^6}\\\\\rightarrow~a_{11}~=~a_1~.~q^{11-1}~~~\rightarrow~~~\boxed{a_{11}~=~a_1~.~q^{10}}$

Vamos agora montar as expressões:

$a_5+a_9:~~~(a_1~.~q^4)~+~(a_1~.~q^8)\\\\\\\boxed{a_5+a_9:~~~a_1~.~(q^4~+~q^8)~=~40}\\\\\\\\a_7+a_{11}:~~~(a_1~.~q^6)~+~(a_1~.~q^{10})\\\\\\\boxed{a_7+a_{11}:~~~a_1~.~(q^6~+~q^{10})~=~360}$

Dividindo a segunda equação pela primeira:

$\frac{a_1~.~(q^6~+~q^{10})}{a_1~.~(q^4~+~q^8)}~=~\frac{360}{40}\\\\\\\frac{q^6~+~q^{10}}{q^4~+~q^8}~=~9\\\\\\\frac{q^2~.~(q^4~+~q^8)}{(q^4~+~q^8)}~=~9\\\\\\q^2~=~9\\\\\\q~=~\pm\sqrt{9}\\\\\\\boxed{q~=~\pm3}$

A razão da PG possui dois possíveis valores: +3 e -3