Respostas
Resposta:
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Resposta:
2)
a_{n}=a_{1}.q^{(n-1)}\\\\Para~a_{5}=810\\\\a_{5}=a_{1}.q^{(5-1)}=a_{1}.q^{4}=810\\\\Para~a_{3}=90\\\\a_{3}=a_{1}.q^{(3-1)}=a_{1}.q^{2}=90\\\\\frac{a_{5}}{a_{3}} =\frac{a_{1}.q^{4}}{a_{1}.q^{2}} =\frac{810}{90}\\\\q^{2}=9 \Rightarrow q=\pm3$, como a PG \' e crescente$~q>0 \Rightarrow q=3\\\\a_{5}=a_{1}.q^{4}=810 \Rightarrow a_{1}.3^{4}=810 \Rightarrow a_{1}=10\\\\a_{7}=10.3^{(7-1)}=10.3^{6}=7.290
3)
a_{n}=a_{1}.q^{(n-1)}\\\\Para~a_{3}=16\\\\a_{3}=a_{1}.q^{(3-1)}=a_{1}.q^{2}=16\\\\Para~a_{6}=1024\\\\a_{6}=a_{1}.q^{(6-1)}=a_{1}.q^{5}=1024\\\\\frac{a_{6}}{a_{3}} =\frac{a_{1}.q^{5}}{a_{1}.q^{2}} =\frac{1024}{16}\\\\q^{3}=64 \Rightarrow q=4\\\\a_{3}=a_{1}.q^{2}=16 \Rightarrow a_{1}.4^{2}=16 \Rightarrow a_{1}=1\\\\a_{8}=1.4^{(8-1)}=1.4^{7}=16.384
4)
(3,6,12,...)\\\\q=\frac{12}{6}=2\\\\a_{1}=3\\\\S_{n}=\frac{a_{1}(q^{n}-1)}{q-1} \\\\S_{8}=\frac{3(2^{8}-1)}{2-1}=\frac{3(256-1)}{1}=3(255)=765
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