Pertanyaan

Dalam suatu barisan geometri, U 1 + U 3 + U 5 = s dan U 2 + U 4 + U 6 = t . Maka U 3 adalah ....

Dalam suatu barisan geometri, $U_{1} + U_{3} + U_{5} = s$ dan $U_{2} + U_{4} + U_{6} = t$ . Maka $U_{3}$ adalah ....

$begin mathsize 14px style fraction numerator st to the power of 4 over denominator straight s to the power of 4 plus straight s squared space straight t squared plus straight t to the power of 4 end fraction end style$
$begin mathsize 14px style fraction numerator straight s squared space straight t cubed over denominator straight s to the power of 4 plus straight s squared space straight t squared plus straight t to the power of 4 end fraction end style$
$begin mathsize 14px style fraction numerator straight s cubed space straight t squared over denominator straight s to the power of 4 plus straight s squared space straight t squared plus straight t to the power of 4 end fraction end style$
$begin mathsize 14px style fraction numerator straight s to the power of 4 space straight t over denominator straight s to the power of 4 plus straight s squared space straight t squared plus straight t to the power of 4 end fraction end style$
$begin mathsize 14px style fraction numerator straight s to the power of 5 over denominator straight s to the power of 4 plus straight s squared space straight t squared plus straight t to the power of 4 end fraction end style$

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Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

Pembahasan

Ingat bahwa pada barisan geometri berlaku bahwa Sehingga dan Sehingga Kemudian, perhatikan bahwa Sehingga Jadi, jawaban yang tepat adalah C.

Ingat bahwa pada barisan geometri berlaku bahwa

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Sehingga

begin mathsize 14px style straight U subscript 1 plus straight U subscript 3 plus straight U subscript 5 equals straight s straight a plus ar squared plus ar to the power of 4 equals straight s end style

dan

begin mathsize 14px style straight U subscript 2 plus straight U subscript 4 plus straight U subscript 6 equals straight t ar plus ar cubed plus ar to the power of 5 equals straight t end style

Sehingga

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator ar plus ar cubed plus ar to the power of 5 over denominator straight a plus ar squared plus ar to the power of 4 end fraction end cell equals cell straight t over straight s end cell row cell fraction numerator straight r left parenthesis straight a plus ar squared plus ar to the power of 4 right parenthesis over denominator straight a plus ar squared plus ar to the power of 4 end fraction end cell equals cell straight t over straight s end cell row straight r equals cell straight t over straight s end cell end table end style$

Kemudian, perhatikan bahwa

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight a plus ar squared plus ar to the power of 4 end cell equals straight s row cell straight a plus straight a open parentheses straight t over straight s close parentheses squared plus straight a open parentheses straight t over straight s close parentheses to the power of 4 end cell equals straight s row cell straight a plus straight a space. space straight t squared over straight s squared plus straight a space. space straight t to the power of 4 over straight s to the power of 4 end cell equals straight s row cell straight s to the power of 4 space open parentheses straight a plus straight a space. space straight t squared over straight s squared plus straight a space. space straight t to the power of 4 over straight s to the power of 4 close parentheses end cell equals cell straight s to the power of 4 left parenthesis straight s right parenthesis end cell row cell as to the power of 4 plus as squared space straight t squared plus at to the power of 4 end cell equals cell straight s to the power of 5 end cell row cell straight a left parenthesis straight s to the power of 4 plus straight s squared space straight t squared plus straight t to the power of 4 right parenthesis end cell equals cell straight s to the power of 5 end cell row straight a equals cell fraction numerator straight s to the power of 5 over denominator straight s to the power of 4 plus straight s squared straight t squared plus straight t to the power of 4 end fraction end cell end table end style$

Sehingga

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell straight U subscript 3 end cell equals cell ar squared end cell row blank equals cell fraction numerator straight s to the power of 5 over denominator straight s to the power of 4 plus straight s squared space straight t squared plus straight t to the power of 4 end fraction. space open parentheses straight t over straight s close parentheses squared end cell row blank equals cell fraction numerator straight s to the power of 5 over denominator straight s to the power of 4 plus straight s squared space straight t squared plus straight t to the power of 4 end fraction. space straight t squared over straight s squared end cell row blank equals cell fraction numerator straight s cubed space straight t squared over denominator straight s to the power of 4 plus straight s squared space straight t squared plus straight t to the power of 4 end fraction end cell end table end style$

Jadi, jawaban yang tepat adalah C.

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