Question

1) qual é o 6° termo da P.G. (-240, -120, -60,..)

2) Em uma P.G., crescente, o 1° termo é √2 e o 3° termo é  5√2. Determine:

a) o 2° termo.  b) a razão da P.G.  c) o 7° termo.

Answer 1

1) $P.G(-240,-120,-60,...)$

$a_{1}=-240$
$a_{2}=-120$

$q = a_{2}/a_{1}=-120/(-240)=1/2$

$a_{n}=a_{1}*q^{(n-1)}$
$a_{6}=a_{1}*q^{(6-1)}$
$a_{6}=a_{1}*q^{5}$
$a_{6}=(-240)*(1/2)^{5}$
$a_{6}=(-240)*(1/32)$
$a_{6}=-240/32$
$a_{6}=-15/2$

2)

a)

$q = a_{2} / a_{1}$
$q = a_{3} / a_{2}$

$q = q$
$a_{2} / a_{1} = a_{3} / a_{2}$
$a_{2}*a_{2}=a_{1}*a_{3}$
$(a_{2})^{2}=a_{1}*a_{3}$
$(a_{2})^{2}=\sqrt{2}*5\sqrt{2}$
$(a_{2})^{2}=5* \sqrt{2} * \sqrt{2}$
$(a_{2})^{2}=5*2$
$(a_{2})^{2}=10$
$a_{2}= \sqrt{10}$

b)

$a_{1}= \sqrt{2}$
$a_{2}= \sqrt{10}$

$q = a_{2}/a_{1}= \sqrt{10}/\sqrt{2}=\sqrt{10/2}= \sqrt{5}$

c)

$a_{n}=a_{1}*q^{(n-1)}$
$a_{7}=a_{1}*q^{(7-1)}$
$a_{7}=a_{1}*q^{6}$
$a_{7}=\sqrt{2}*(\sqrt{5})^{6}$
$a_{7}=\sqrt{2}*5^{6/2}$
$a_{7}=\sqrt{2}*5^{3}$
$a_{7}=\sqrt{2}*125$
$a_{7}=125\sqrt{2}$