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Diketahui suku ke-3 dan suku ke-6 suatu deret geometri berturut-turut adalah 48 dan 384. jumlah lima suku pertama dari deret tersebut adalah ...

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$R u m u s space s u k e space k e minus n space y a i t u U subscript n equals space a r to the power of n minus 1 end exponent r u m u s space j u m l a h space s a m p a i space s u k u space k e minus n space a d a l a h S subscript n equals fraction numerator a left parenthesis r to the power of n minus 1 right parenthesis over denominator r minus 1 end fraction colon space r space greater than space 1 D a n space u n t u k space m e n c a r i space r a s i o space a d a l a h space r to the power of m minus n end exponent equals U subscript m over U subscript n m a k a comma r to the power of left parenthesis 6 minus 3 right parenthesis end exponent equals space U subscript 6 over U subscript 3 equals 384 over 48 equals 8 r cubed equals 8 r equals 2 U subscript n equals space a r to the power of n minus 1 end exponent a equals U subscript n over r to the power of n minus 1 end exponent equals U subscript 3 over r to the power of 3 minus 1 end exponent equals 48 over 2 squared equals 12 S e h i n g g a colon S subscript n equals fraction numerator a left parenthesis r to the power of n minus 1 right parenthesis over denominator r minus 1 end fraction semicolon space r greater than space 1 S subscript 5 equals fraction numerator 12 left parenthesis 2 to the power of 5 minus 1 right parenthesis over denominator 2 minus 1 end fraction equals space 372 J u m l a h space s u k u space p e r t a m a n y a space y a i t u space 372$