Question

invente uma situação de progressão geométrica , e depois resolva ela.

Pode ser uma PA simples.
Alguem me ajuda..​

Answer 1

$\Large\boxed{\begin{array}{l}\sf descubra\,quantos\,termos\,possui\\\sf a\,PG\,(2,4,8,16\dotsc512).\\\underline{\rm soluc_{\!\!,}\tilde ao\!:}\\\sf a_1=2\\\sf q=\dfrac{4}{2}=2\\\sf n=?\\\sf a_n=512.\\\sf a_n=a_1\cdot q^{n-1}\\\sf 512=2\cdot2^{n-1}\\\sf lembrando\,que\,a^m\cdot a^n=a^{m+n}\,temos;\\\sf 2\cdot2^{n-1}=512\\\sf 2^{\backslash\!\!\!1+n-\backslash\!\!\!1}=512\\\sf 2^n=512\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Decompondo\,512}\\\underline{\rm em\,fatores\,primos\,temos:}\\\begin{array}{c|c}\sf512&\sf2\\\sf256&\sf2\\\sf128&\sf2\\\sf64&\sf2\\\sf32&\sf2\\\sf16&\sf2\\\sf8&\sf2\\\sf4&\sf2\\\sf2&\sf2\\\sf1\end{array}\\\sf 512=2^9.\\\sf substituindo\,na\,equac_{\!\!,}\tilde ao\,teremos:\\\sf 2^n=2^9\\\sf n=9\checkmark\\\tt Portanto\,esta\,PG\,possui\,9\,termos.\end{array}}$