P.G qual a ordem do termo igual a 192 na P.G. (3,6,12,...)
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Vamos lá:
![a_{n} = 192 \\ a_{1} = 3 \\ a_{2} = 6 \\ a_{n} = a_{1} . q^{n-1} \\ a_{2} = a_{1} . q^{1} \\ 6 = 3 . q \\ q = 2 \\ a_{n} = a_{1}.2^{n-1} \\ 192 = 3 . 2^{n-1} \\ 2^{n-1} = 64 \\ 2^{n-1} = 2^{6} \\ n-1 = 6 \\ n = 7 \\
a_{7} = 192 a_{n} = 192 \\ a_{1} = 3 \\ a_{2} = 6 \\ a_{n} = a_{1} . q^{n-1} \\ a_{2} = a_{1} . q^{1} \\ 6 = 3 . q \\ q = 2 \\ a_{n} = a_{1}.2^{n-1} \\ 192 = 3 . 2^{n-1} \\ 2^{n-1} = 64 \\ 2^{n-1} = 2^{6} \\ n-1 = 6 \\ n = 7 \\
a_{7} = 192](https://tex.z-dn.net/?f=a_%7Bn%7D+%3D+192+%5C%5C+a_%7B1%7D+%3D+3+%5C%5C+a_%7B2%7D+%3D+6+%5C%5C+a_%7Bn%7D+%3D+a_%7B1%7D+.+q%5E%7Bn-1%7D+%5C%5C+a_%7B2%7D+%3D+a_%7B1%7D+.+q%5E%7B1%7D+%5C%5C+6+%3D+3+.+q+%5C%5C+q+%3D+2+%5C%5C+a_%7Bn%7D+%3D+a_%7B1%7D.2%5E%7Bn-1%7D+++%5C%5C+192+%3D+3+.+2%5E%7Bn-1%7D+%5C%5C+2%5E%7Bn-1%7D+%3D+64+%5C%5C+2%5E%7Bn-1%7D+%3D+2%5E%7B6%7D+%5C%5C+n-1+%3D+6+%5C%5C+n+%3D+7+%5C%5C%0Aa_%7B7%7D+%3D+192)
Espero ter ajudado, qualquer dúvida, contate-me!
Espero ter ajudado, qualquer dúvida, contate-me!
Respondido por
24
P.G qual a ordem do termo igual a 192 na P.G. (3,6,12,...)
an = a1.q^(n-1)
3.2^(n-1) = 192
2^(n-1) = 192/3
2^(n-1) = 64
2^(n-1) = 2^6
n - 1 = 6
n = 6 + 1
n =7
an = a1.q^(n-1)
3.2^(n-1) = 192
2^(n-1) = 192/3
2^(n-1) = 64
2^(n-1) = 2^6
n - 1 = 6
n = 6 + 1
n =7
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