Question

alguem poderia me ajudar com essa PG?

Answer 1

$a_5-a_3=a_1.q^4-a_1.q^2=a_1(q^4-q^2)=16 \Rightarrow \boxed{a_1=\frac{16}{q^4-q^2}}\\ \\ \\ a_9-a_7=a_1.q^8-a_1.q^6=a_1(q^8-q^6)=1296 \Rightarrow \boxed{a_1=\frac{1296}{q^8-q^6}}\\ \\ \\ \frac{16}{q^4-q^2}=\frac{1296}{q^8-q^6}\\ \\ \frac{1}{q^2(q^2-1)}=\frac{81}{q^6(q^2-1)}\\ \\ \frac{1}{q^2}=\frac{81}{q^6}\\ \\ q^6=81q^2\\ \\ \frac{q^6}{q^2}=81\\ \\ q^4=81\\ \\ \boxed{\boxed{q=3}}$

$\\ \\ a_1=\frac{16}{q^4-q^2}\\ \\ a_1=\frac{16}{3^4-3^2}\\ \\ a_1=\frac{16}{81-9}\\ \\ a_1=\frac{16}{72}\\ \\ \boxed{\boxed{a_1=\frac{2}{9}}}$