Question

Preciso da conta. 50 pontos
Dados "a1" e "q" determine os quatro primeiros termos da PG abaixo:
a1: -9         q: 4
a1: -60        q: -1/2
a1: 1/5        q: 7/3

Answer 1

$P.G\ em\ geral \to (a_1,a_1*q,a_1*q^2,...,a_1*q^{n-1})\\\\P.G(1)\to(-9,-36,-144,-576)\\\\P.G(2)\to(-60, 30, -15, \frac{15}{2})\\\\P.G(3) \to ( \frac{1}{5}, \frac{7}{15}, \frac{49}{45}, \frac{343}{135} )$

Espero ter ajudado. :))

Answer 2

Termo geral : $An = a1.q^{n-1}$

primeira PG :

$A_2 = -9.4^{2-1}\to A_2 = -9.4\to A_2 = -36 \\ A_3 = -9.4^{3-1}\to A_3 = -9.4^2 \to A_3 = -144 \\ A_4 = -9.4^{4-1}\to A_4 = -9.4^3 \to A_4 = -576 \\ (-9;-36;-144;-576)$

Segunda PG :

$A_2 = -60.(- \frac{1}{2})^{2-1} \to A_2 = -60.- \frac{1}{2} \to A_2 = 30 \\ A_3 = -60.(- \frac{1}{2})^{3 - 1} \to A_3 = -60.(- \frac{1}{2})^2 \to A_3 = 15 \\ A_4 = -60.(- \frac{1}{2})^{4-1}\to A_4 = -60.(- \frac{1}{2}) ^3 \to A_4 = \frac{15}{2} \\ (-60;30;15;\frac{15}{2} )$

Terceira PG :

$A_2 = \frac{1}{5} . \frac{7}{5} ^{2-1}\to A_2 = \frac{1}{5} . \frac{7}{5} \to A_2 = \frac{7}{25} \\ \\ A_3 = \frac{1}{5} . \frac{7}{5} ^{3-1}\to A_3 = \frac{1}{5} . \frac{49}{25} \to A_3 = \frac{49}{125} \\ \\ A_4 = \frac{1}{5} . \frac{7}{5} ^{4-1} \to A_4 = \frac{1}{5} . \frac{323}{125}\to A_4 = \frac{323}{625} \\ \\ (\frac{1}{5}; \frac{7}{25} ; \frac{49}{125}; \frac{323}{625})$