Considere a PG de razão Q = ⁷√₃ e A15 = 5 e determine A1.
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Pela fórmula do termo geral da P.G., vem:
![\boxed{a _{n}=a _{1}*q ^{n-1}}\\\\
5=a _{1}*( \sqrt[7]{3}) ^{15-1}\\
5=a _{1}*( \sqrt[7]{3}) ^{14}\\
5=a _{1}*(3^{ \frac{1}{7} }) ^{14}\\
5=a _{1}*3 ^{2}\\
9a_{1}=5\\\\
\boxed{a _{1}= \frac{5}{9}}](https://tex.z-dn.net/?f=%5Cboxed%7Ba+_%7Bn%7D%3Da+_%7B1%7D%2Aq+%5E%7Bn-1%7D%7D%5C%5C%5C%5C%0A5%3Da+_%7B1%7D%2A%28+%5Csqrt%5B7%5D%7B3%7D%29+%5E%7B15-1%7D%5C%5C%0A5%3Da+_%7B1%7D%2A%28+%5Csqrt%5B7%5D%7B3%7D%29+%5E%7B14%7D%5C%5C%0A5%3Da+_%7B1%7D%2A%283%5E%7B+%5Cfrac%7B1%7D%7B7%7D+%7D%29+%5E%7B14%7D%5C%5C%0A5%3Da+_%7B1%7D%2A3+%5E%7B2%7D%5C%5C%0A9a_%7B1%7D%3D5%5C%5C%5C%5C%0A%5Cboxed%7Ba+_%7B1%7D%3D+%5Cfrac%7B5%7D%7B9%7D%7D+++++++++++++++++ "\boxed{a _{n}=a _{1}*q ^{n-1}}\\\\
5=a _{1}*( \sqrt[7]{3}) ^{15-1}\\
5=a _{1}*( \sqrt[7]{3}) ^{14}\\
5=a _{1}*(3^{ \frac{1}{7} }) ^{14}\\
5=a _{1}*3 ^{2}\\
9a_{1}=5\\\\
\boxed{a _{1}= \frac{5}{9}}")
Espero ter ajudado e tenha bons estudos!!!
Espero ter ajudado e tenha bons estudos!!!
jvccjoaovitor:
Ajudou sim, muito obrigado!!
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