Question

a) determine o 12° termos da PG (17,34,...)

b) determine o 13° termo da PG (11,22,...)

c) quantos termos possui a PG (1,3,9,...,2187)

d) interpole 5 meios geométricos entre 6 e 384

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Answer 1

$> resolucao \\ \\ \\ a)12 \: termo \: da \: pg \: (17.34...) \\ \\ q = \frac{a2}{a1} \\ \\ q = \frac{34}{17} \\ \\ q = 2 \\ \\ an = a1 \times q {}^{n - 1} \\ an = 17 \times 2 {}^{12 - 1 } \\ an = 17 \times 2 {}^{11} \\ an = 17 \times 2048 \\ an = 34816 \\ \\ \\ b) \: 13 \: termo \: da \: pg \: (11 \: . \: 22...) \\ \\ q = \frac{a2}{a1} \\ \\ q = \frac{22}{11} \\ \\ q = 2 \\ \\ an = a1 \times q {}^{n - 1} \\ an = 11 \times 2 {}^{13 - 1} \\ an = 11 \times 2 {}^{12} \\ an = 11 \times 4096 \\ an = 45056 \\ \\ \\ c) \: numero \: de \: termos \: d a \: pg \: (1.3.9.....2187) \\ \\ \\ an = a1 \times q {}^{n - 1} \\ 2187 = 1 \times 3 {}^{n - 1} \\ \frac{2187}{1} = 3 {}^{n - 1} \\ 2187 = 3 {}^{n - 1} \\ 3 {}^{7} = 3 {}^{n - 1} \\ n - 1 = 7 \\ n = 7 + 1 \\ n = 8 \\ \\ \\ d) \: interpolando \: 5 \: meios \: entre \: 6 \: e \: 384 \\ \\ \\ an = a1 \times q {}^{n - 1} \\ 384 = 6 \times q {}^{7 - 1} \\ \frac{384}{6} = q {}^{6} \\ 64 = q {}^{6} \\ 2 {}^{6} = q {}^{6} \\ q = 2 \\ \\ \\ pg \: (6.12.24.48.96.192.384...) \\ \\ \\ \\ > > > > > > >$