Iklan

Iklan

Pertanyaan

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

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

Iklan

A. Rizky

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Iklan

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

 

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

1

Iklan

Iklan

Pertanyaan serupa

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

1

2.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia