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Pertanyaan

Jika a + 1, a - 3, 2 membentuk barisan geometri, maka jumlah 11 suku pertama yang mungkin adalah ....

  1. 2

  2. 4

  3. 6

  4. 7

  5. 9

A. Rizky

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Pembahasan

Ingat space bahwa space pada space barisan space geometri space berlaku  straight r equals straight u subscript 2 over straight u subscript 1 equals straight u subscript 3 over straight u subscript 2    Maka  fraction numerator straight a minus 3 over denominator straight a plus 1 end fraction equals fraction numerator 2 over denominator straight a minus 3 end fraction  open parentheses straight a minus 3 close parentheses squared equals 2 open parentheses straight a plus 1 close parentheses  straight a squared minus 6 straight a plus 9 equals 2 straight a plus 2  straight a squared minus 8 straight a plus 7 equals 0  open parentheses straight a minus 1 close parentheses open parentheses straight a minus 7 close parentheses equals 0    Sehingga  straight a minus 1 equals 0 rightwards arrow box enclose straight a equals 1 end enclose  straight a minus 7 equals 0 rightwards arrow box enclose straight a equals 7 end enclose    Jika space straight a equals 1 comma space maka  straight u subscript 1 equals straight a plus 1 equals 1 plus 1 equals box enclose 2  straight r equals fraction numerator straight a minus 3 over denominator straight a plus 1 end fraction equals fraction numerator 1 minus 3 over denominator 1 plus 1 end fraction equals fraction numerator negative 2 over denominator 2 end fraction equals box enclose negative 1 end enclose    Jika space straight a equals 7 comma space maka  straight u subscript 1 equals straight a plus 1 equals 7 plus 1 equals box enclose 8  straight r equals fraction numerator straight a minus 3 over denominator straight a plus 1 end fraction equals fraction numerator 7 minus 3 over denominator 7 plus 1 end fraction equals 4 over 8 equals box enclose 1 half end enclose    Jika space straight r equals 1 comma space maka  straight S subscript 11 equals fraction numerator 2 open parentheses open parentheses negative 1 close parentheses to the power of 11 minus 1 close parentheses over denominator open parentheses negative 1 close parentheses minus 1 end fraction equals fraction numerator 2 open parentheses negative 1 minus 1 close parentheses over denominator negative 1 minus 1 end fraction equals fraction numerator 2 open parentheses negative 2 close parentheses over denominator negative 2 end fraction equals box enclose 2    Jika space straight r equals 7 comma space maka space  straight S subscript 11 equals fraction numerator 8 open parentheses open parentheses 1 half close parentheses to the power of 11 minus 1 close parentheses over denominator open parentheses 1 half close parentheses minus 1 end fraction equals fraction numerator 8 open parentheses 1 over 2048 minus 1 close parentheses over denominator negative 1 half end fraction equals fraction numerator 1 over 256 minus 8 over denominator negative 1 half end fraction equals fraction numerator negative 2047 over 256 over denominator negative 1 half end fraction equals box enclose 2047 over 128 end enclose

 

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