Em uma progressão geométrica crescente, o quinto termo é 5 e o décimo termo é 15625. Qual é o valor do 8º termo dessa progressão?
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Ae mano,
como trata-se de uma P.G. crescente, temos que a razão deve ser positiva, então acompanhe passo a passo:
![P.G.\underbrace{(a_5,a_6,a_7,a_8,a_9,a_{10})}\\
~~~~~~~~~~~~~~~6~termos\\\\
a_n=a_1\cdot q^{n-1}\\
15.625=5\cdot q^{6-1}\\\\
q^5= \dfrac{15.625}{5}\\\\
q^5=3.125\\
q=\pm\sqrt[5]{3.125}\\\\
\boxed{q=\pm5}\\\\
a_n=a_1\cdot q^{n-1}\\
15.625=a_1\cdot 5^{10}\\
a_1\cdot5^{10}=5^6\\\\
a_1= \dfrac{5^6}{5^{10}}\\\\
a_1=5^{-4}\\\\
a_1= \dfrac{1}{5^4}\\\\
\boxed{a_1= \dfrac{1}{625}}\\\\\\
a_8=a_1\cdot q^7\\\\
a_8= \dfrac{1}{625}\cdot5^7\\\\
a_8= \dfrac{1}{625}\cdot78.125\\\\\\
\huge\boxed{\boxed{a_8=125}}}](https://tex.z-dn.net/?f=P.G.%5Cunderbrace%7B%28a_5%2Ca_6%2Ca_7%2Ca_8%2Ca_9%2Ca_%7B10%7D%29%7D%5C%5C%0A%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E6%7Etermos%5C%5C%5C%5C%0Aa_n%3Da_1%5Ccdot+q%5E%7Bn-1%7D%5C%5C%0A15.625%3D5%5Ccdot+q%5E%7B6-1%7D%5C%5C%5C%5C%0Aq%5E5%3D+%5Cdfrac%7B15.625%7D%7B5%7D%5C%5C%5C%5C%0Aq%5E5%3D3.125%5C%5C%0A+q%3D%5Cpm%5Csqrt%5B5%5D%7B3.125%7D%5C%5C%5C%5C%0A%5Cboxed%7Bq%3D%5Cpm5%7D%5C%5C%5C%5C%0Aa_n%3Da_1%5Ccdot+q%5E%7Bn-1%7D%5C%5C%0A15.625%3Da_1%5Ccdot+5%5E%7B10%7D%5C%5C%0Aa_1%5Ccdot5%5E%7B10%7D%3D5%5E6%5C%5C%5C%5C%0Aa_1%3D+%5Cdfrac%7B5%5E6%7D%7B5%5E%7B10%7D%7D%5C%5C%5C%5C%0Aa_1%3D5%5E%7B-4%7D%5C%5C%5C%5C%0Aa_1%3D+%5Cdfrac%7B1%7D%7B5%5E4%7D%5C%5C%5C%5C%0A%5Cboxed%7Ba_1%3D+%5Cdfrac%7B1%7D%7B625%7D%7D%5C%5C%5C%5C%5C%5C+++++%0Aa_8%3Da_1%5Ccdot+q%5E7%5C%5C%5C%5C%0Aa_8%3D+%5Cdfrac%7B1%7D%7B625%7D%5Ccdot5%5E7%5C%5C%5C%5C%0Aa_8%3D+%5Cdfrac%7B1%7D%7B625%7D%5Ccdot78.125%5C%5C%5C%5C%5C%5C%0A%5Chuge%5Cboxed%7B%5Cboxed%7Ba_8%3D125%7D%7D%7D "P.G.\underbrace{(a_5,a_6,a_7,a_8,a_9,a_{10})}\\
~~~~~~~~~~~~~~~6~termos\\\\
a_n=a_1\cdot q^{n-1}\\
15.625=5\cdot q^{6-1}\\\\
q^5= \dfrac{15.625}{5}\\\\
q^5=3.125\\
q=\pm\sqrt[5]{3.125}\\\\
\boxed{q=\pm5}\\\\
a_n=a_1\cdot q^{n-1}\\
15.625=a_1\cdot 5^{10}\\
a_1\cdot5^{10}=5^6\\\\
a_1= \dfrac{5^6}{5^{10}}\\\\
a_1=5^{-4}\\\\
a_1= \dfrac{1}{5^4}\\\\
\boxed{a_1= \dfrac{1}{625}}\\\\\\
a_8=a_1\cdot q^7\\\\
a_8= \dfrac{1}{625}\cdot5^7\\\\
a_8= \dfrac{1}{625}\cdot78.125\\\\\\
\huge\boxed{\boxed{a_8=125}}}")
ESPERO TER AJUDADO E TENHA ÓTIMOS ESTUDOS ;D
como trata-se de uma P.G. crescente, temos que a razão deve ser positiva, então acompanhe passo a passo:
ESPERO TER AJUDADO E TENHA ÓTIMOS ESTUDOS ;D
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