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Pertanyaan

Jika u 1 ,u 2 ,u 3 , ... adalah barisan geometri yang memenuhi u 3 - u 6 = x, dan u 2 - u 4 = y, maka x/y = ...

Jika u1,u2,u3, ... adalah barisan geometri yang memenuhi u- u= x, dan u- u= y, maka x/y = ...

  1. begin mathsize 14px style bevelled fraction numerator open parentheses r cubed minus r squared minus r close parentheses over denominator open parentheses r minus 1 close parentheses end fraction end style

  2. bevelled fraction numerator open parentheses size 14px r to the power of size 14px 3 size 14px minus size 14px r to the power of size 14px 2 size 14px plus size 14px r close parentheses over denominator open parentheses size 14px r size 14px minus size 14px 1 close parentheses end fraction

  3. bevelled fraction numerator open parentheses size 14px r to the power of size 14px 3 size 14px plus size 14px r to the power of size 14px 2 size 14px plus size 14px r close parentheses over denominator open parentheses size 14px r size 14px plus size 14px 1 close parentheses end fraction

  4. bevelled fraction numerator open parentheses size 14px r to the power of size 14px 3 size 14px plus size 14px r to the power of size 14px 2 size 14px minus size 14px r close parentheses over denominator open parentheses size 14px r size 14px minus size 14px 1 close parentheses end fraction

  5. bevelled fraction numerator open parentheses size 14px r to the power of size 14px 3 size 14px minus size 14px r to the power of size 14px 2 size 14px plus size 14px r close parentheses over denominator open parentheses size 14px r size 14px plus size 14px 1 close parentheses end fraction

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A. Rizky

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

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Pembahasan

begin mathsize 14px style text i) Perhatikan rumus dari barisan geometri dan informasi yang diberikan end text  u subscript n equals a r to the power of n minus 1 end exponent  u subscript n equals u subscript k r to the power of n minus k end exponent  S subscript n equals fraction numerator a open parentheses r to the power of n minus 1 close parentheses over denominator r minus 1 end fraction  text Maka: end text  x over y equals fraction numerator u subscript 3 minus u subscript 6 over denominator u subscript 2 minus u subscript 4 end fraction equals fraction numerator a r squared minus a r to the power of 5 over denominator a r minus a r cubed end fraction equals fraction numerator a r open parentheses r minus r to the power of 4 close parentheses over denominator a r open parentheses 1 minus r squared close parentheses end fraction equals fraction numerator r minus r to the power of 4 over denominator 1 minus r squared end fraction  equals fraction numerator r open parentheses 1 minus r cubed close parentheses over denominator open parentheses 1 minus r close parentheses open parentheses 1 plus r close parentheses end fraction equals fraction numerator r open parentheses 1 minus r close parentheses open parentheses 1 plus r plus r squared close parentheses over denominator open parentheses 1 minus r close parentheses open parentheses 1 plus r close parentheses end fraction  equals fraction numerator r open parentheses 1 plus r plus r squared close parentheses over denominator open parentheses 1 plus r close parentheses end fraction equals fraction numerator r plus r squared plus r cubed over denominator 1 plus r end fraction  end style

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Pertanyaan serupa

Diketahui selisih jumlah enam suku pertama dan jumlah ketiga suku pertama suatu deret geometri dengan rasio positif adalah 168. Jika suku ke 4 deret tersebut adalah 24,maka suku kedua deret geometri t...

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