Diketahui suatu deret geometri mempunyai suku-suku positif. Suku ke-3 = 36 dan suku ke-5 = 324. Jumlah 6 suku pertama adalah ....

Pertanyaan

1452
1454
1456
1458
1460

R. Diah

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Pembahasan

straight U subscript 3 equals ar to the power of 3 minus 1 end exponent 36 equals ar squared straight U subscript 5 equals ar to the power of 5 minus 1 end exponent 324 equals ar to the power of 4

Dengan membandingkan U₅ dan U₃

$fraction numerator straight U subscript 5 equals ar to the power of 4 over denominator straight U subscript 3 equals ar squared end fraction fraction numerator 324 equals ar to the power of 4 over denominator 36 equals ar squared end fraction 9 equals straight r squared straight r equals 3$

Substitusi nilai r ke dalam U₃

straight U subscript 3 equals ar squared 36 equals straight a left parenthesis 9 right parenthesis straight a equals 4

Rumus S_n

$straight S subscript straight n equals fraction numerator straight a left parenthesis straight r to the power of straight n minus 1 right parenthesis over denominator straight r minus 1 end fraction$

Jadi, Jumlah enam suku pertama :

$straight S subscript 6 equals fraction numerator 4 left parenthesis 3 to the power of 6 minus 1 right parenthesis over denominator 3 minus 1 end fraction straight S subscript 6 equals 2 left parenthesis 729 minus 1 right parenthesis straight S subscript 6 equals 2 left parenthesis 728 right parenthesis straight S subscript 6 equals 1456$