Matematika tentukan unsur yang di minta pada barisan geometri berikut
a. a=-3,U4=1/9:r=......

b. U3=8,U4=32:a=........ ...

c U2=250,U4=6.250:a=......

d. U2=12,U5=-324:r=......

e. k-2,k-6,2k+39,......:k=

tentukan unsur yang di minta pada barisan geometri berikut
a. a=-3,U4=1/9:r=......

b. U3=8,U4=32:a=........ ...

c U2=250,U4=6.250:a=......

d. U2=12,U5=-324:r=......

e. k-2,k-6,2k+39,......:k=

2. a. a = -3, U₄ = 1/9

[tex]un = ar {}^{n - 1} \\ u4 = - 3.r {}^{4 - 1} \\ \frac{1}{9} = - 3.r {}^{3} [/tex]

1/9. -⅓ = r³

-1/27 = r³

[tex]r = \sqrt[3]{ - \frac{1}{27} } = - \frac{1}{3} [/tex]

b. U₃ = 8 ------> ar² = 8

U₄ = 32 -----> ar³ = 32

[tex] \frac{ar {}^{3} }{ar {}^{2} } = \frac{32}{8} \\ r = 4[/tex]

ar² = 8

a.4² = 8

16a = 8

a = 8/16 = ½

c. U₂ = 250 ------> ar = 250

U₄ = 6250 ------> ar³ = 6250

[tex] \frac{ar {}^{3} }{ar } = \frac{6250}{250} \\ r {}^{2} = 25 \\ r = \sqrt{25} = 5[/tex]

ar = 250

a.5 = 250

a = 250/5 = 50

d. U₂ = 12 --------> ar = 12

U₅ = -324 --------> ar⁴ = -324

[tex] \frac{ar {}^{4} }{ar} = \frac{ - 324}{12} \\ r {}^{3} = - 27 \\ r = \sqrt[3]{ - 27} = - 3[/tex]

e. k - 2, k - 6, 2k + 3

Geometri

Rasio = Rasio

[tex] \frac{u2}{u1} = \frac{u3}{u2} \\ \frac{k - 6}{k - 2} = \frac{2k + 3}{k - 6} [/tex]

( k - 6 ) ( k - 6 ) = ( 2k + 3 ) ( k - 2 )

k² - 12k + 36 = 2k² - k - 6

2k² - k² - k + 12k - 6 - 36 = 0

k² + 11k - 42 = 0

( k + 14 ) ( k - 3 ) = 0

k = -14, k = 3

[answer.2.content]