Question

qual o numero de termo da PG (9,3,...,1/81

Answer 1

$P.G.(9,3,...,\frac{1}{81})\\ a_1=9\\ q=\frac{a_2}{a_1}=\frac{3}{9}=\frac{1}{3}\\\\ a_n=a_1.q^{n-1}\\\\ \frac{1}{81}=9.(\frac{1}{3})^{n-1}\\\\ 9.(\frac{1}{3})^{n-1}=\frac{1}{81}\\\\ (\frac{1}{3})^{n-1}=\frac{1}{81.9}\\\\ (\frac{1}{3})^{n-1}=\frac{1}{729}\\\\ (\frac{1}{3})^{n-1}=\frac{1}{3^6}\\\\ (\frac{1}{3})^{n-1}=(\frac{1}{3})^6\\\\ n-1=6\\ n=6+1\\ n=7$