[tex]Hola.\\
\frac{4}{15} *\frac{5}{2}=\frac{2}{3}\\a_n=a_1*q^ {(n-1)}\\\frac{256}{10935}=\frac{2}{5}*(\frac{2}
{3})^{(n-1)}\\\frac{256*5}{10956}=2*(\frac{2}{3})^{(n-1)}\\\frac{256}{2187}=2*(\frac{2}{3})^{(n-1)}\\\frac{256}{2187*2}=(\frac{2}{3})^{(n-1)}, como:256=2^8~~ e~~ 2187=3^7, então:\\\frac{2^8}{3^3*2}=(\frac{2}{3})^{(n-1)}\\\frac{2^8*2^-1}{3^7}=(\frac{2}{3})^{n-1}\\\frac{2^7}{3^7}=(\frac{2}{3})^{n-1}\\ (\frac{2}{3})^7=(\frac{2}{3})^{n-1},bases~iguais~se~corta:\\7=n-1\\n=7+1\\\boxed{n=8}~oitavo~termo.[/tex]
