Question

progresiei geometrice ​

Answer 1

Răspuns:

Explicație pas cu pas:

a) $b_{2}$=6*2=12

$b_{3}$=6*$2^{2}$ =24

$b_{4}$=$b_{3}$*q=24*2=48

$b_{5}$=$b_{4}$ *q

b) $b_{2}$=$b_{1}$ *q

-10=$b_{1}$ * $\frac{1}{2}$

$b_{1}$=-20

$b_{3}$=$b_{2}$ *q=-$\frac{10}{2}$=-5

$b_{4}$=$b_{3}$ *q=$\frac{-5}{2}$

$b_{5}$=$b_{4}$ *q=$\frac{-5}{4}$

2. a) $b_{1}$=5

$b_{2}$=25

$b_{2}$=$b_{1}$*q => q=25:5=5

x=$b_{3}$=25*5=125

y=$b_{4}$=125*5=625

z=625*5=3125

b) $b_{4}$=$b_{3}$*q => 36=24q =>q=36:24=$\frac{3}{2}$

$b_{3}$=$b_{2}$*q

24=y * $\frac{3}{2}$ => y=48:3=16

$b_{2}$=$b_{1}$*q

16=x*$\frac{3}{2}$ => x=$\frac{32}{3}$

c) 9=1* $q^{2}$ => q=+/- 3

z=3 sau z=-3

y=9*3=27 sau y=-27

z=81*3=163 sau z=-163

3) a) b3=b1 * q^2

b5=b1* q^4

b5 : b3= q^2 => q^2=24:6=4 => q=2

b1=b3 : q^2=6/4=3/2

b)$\left \{ {{b1 *q-b1=4} \atop {b1* q^2 -b1=-8</p><p>}} \right.$ => $\left \{ {{b1(q-1)=4} \atop {b1(q -1)(q+1)=-8</p><p>}} \right.$ => q+1=-8:4=-2, b1=4: (-3)=-4/3

c) $\left \{ {{b1(1+q+q^2)=7} \atop {b1*q(1+q+q^2)=14}} \right.$

prima împărțită la a doua => q=14/7=2

b1(1+2+4)=7 => b1=7/7=1