quantos meios aritmeticos devem ser inseridos entre 7/8 e 14, de forma a se obter uma pa de razão 1/8?
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Se,
![a _{1}= \frac{7}{8}\\\\
a _{n}=14\\\\
r= \frac{1}{8}\\\\
n=? a _{1}= \frac{7}{8}\\\\
a _{n}=14\\\\
r= \frac{1}{8}\\\\
n=?](https://tex.z-dn.net/?f=a+_%7B1%7D%3D++%5Cfrac%7B7%7D%7B8%7D%5C%5C%5C%5C%0Aa++_%7Bn%7D%3D14%5C%5C%5C%5C%0Ar%3D++%5Cfrac%7B1%7D%7B8%7D%5C%5C%5C%5C%0An%3D%3F+)
Pela fórmula do termo geral, vem:
![a _{n}=a _{1}+(n-1)r\\\\ 14= \frac{7}{8}+(n-1) \frac{1}{8}\\\\14= \frac{7}{8}+ \frac{1}{8}n- \frac{1}{8}\\\\ 14= \frac{1}{8}n+ \frac{6}{8}\\\\ 14- \frac{6}{8}= \frac{1}{8}n\\\\ \frac{106}{\not8}= \frac{1}{\not8} n\\\\ n=106~termos a _{n}=a _{1}+(n-1)r\\\\ 14= \frac{7}{8}+(n-1) \frac{1}{8}\\\\14= \frac{7}{8}+ \frac{1}{8}n- \frac{1}{8}\\\\ 14= \frac{1}{8}n+ \frac{6}{8}\\\\ 14- \frac{6}{8}= \frac{1}{8}n\\\\ \frac{106}{\not8}= \frac{1}{\not8} n\\\\ n=106~termos](https://tex.z-dn.net/?f=a+_%7Bn%7D%3Da++_%7B1%7D%2B%28n-1%29r%5C%5C%5C%5C+14%3D++%5Cfrac%7B7%7D%7B8%7D%2B%28n-1%29++%5Cfrac%7B1%7D%7B8%7D%5C%5C%5C%5C14%3D++%5Cfrac%7B7%7D%7B8%7D%2B++%5Cfrac%7B1%7D%7B8%7Dn-++%5Cfrac%7B1%7D%7B8%7D%5C%5C%5C%5C+14%3D++%5Cfrac%7B1%7D%7B8%7Dn%2B++%5Cfrac%7B6%7D%7B8%7D%5C%5C%5C%5C+14-++%5Cfrac%7B6%7D%7B8%7D%3D++%5Cfrac%7B1%7D%7B8%7Dn%5C%5C%5C%5C++%5Cfrac%7B106%7D%7B%5Cnot8%7D%3D++%5Cfrac%7B1%7D%7B%5Cnot8%7D+n%5C%5C%5C%5C+n%3D106%7Etermos)
Se temos 106 termos, subtraímos os extremos e teremos 104 meios.
Espero ter ajudado e tenha ótimos estudos :)
Pela fórmula do termo geral, vem:
Se temos 106 termos, subtraímos os extremos e teremos 104 meios.
Espero ter ajudado e tenha ótimos estudos :)
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