Question

Progressao geometrica

Por favor ajudem com a Letra B ,

Answer 1

$P.G \to \left\{\begin{array}{ccc}( \frac{ \sqrt{3} }{27}, \frac{1}{27}, \frac{ \sqrt{3} }{81},..., \frac{1}{729}) \\\\ Termo\ Geral\to a_n = a_1*q^{n-1}\\\\Onde,\\a_n = u\´ltimo\ termo( \frac{1}{729} )\\a_1 = primeiro\ termo( \frac{ \sqrt{3} }{27} )\\q = raz\~ao( \frac{ \sqrt{3} }{3}) \\n = n\´umero\ de\ termos(?)\end{array}\right$

$Resolu\c{c}\~ao \to \left\{\begin{array}{ccc}a_n = a_1*q^{n-1}&3^{- \frac{n}{2}+ \frac{1}{2} } = 3^{- \frac{7}{2} }\\\\ \frac{1}{729} = \frac{ \sqrt{3} }{27}*( \frac{ \sqrt{3} }{3} )^{n-1}&- \frac{n}{2}+ \frac{1}{2} = - \frac{7}{2} \\\\( \frac{ \sqrt{3} }{3} )^{n-1} = \frac{ \sqrt{3} }{81}&n=8\\\\ ( \frac{ 3^{ \frac{1}{2} } }{3} )^{n-1} = \frac{ 3^{ \frac{1}{2} } }{3^4}\\\\(3^{ \frac{1}{2}-1 })^{n-1} = (3^{ \frac{1}{2}-4 })\\\\(3^{- \frac{1}{2})^{n-1} } = 3^{- \frac{7}{2} }\to \end{array}\right$

Número de termos = 8.

Espero ter ajudado. :))