+39 The fifth term of a geometric sequence of positive numbers is 11 and the eleventh term is 5. What is the eighth term of the sequence? Express your answer in simplest radical form.
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+129899 Let the first term = a1
5th term = a1*r^4 = 11 ⇒ a1 = 11/r^4 (1)
11th term = a1*r^10 = 5 (2)
Subbing (1) into (2) we have
(11/r^4) * r^10 = 5
11r^6 = 5
r^6 = 5/11 take the 6th root of both sides
r = ( 5/11)^(1/6)
a1 = 11/r^4 = 11/ [(5/11)^(1/6)]^4 = 11/ (5/11)^(4/6)
So...the 8th term is
a1*r^7 =
( 11/ (5/11)^(4/6) ) * [(5/11)^(1/6)]^7 =
11 [ (5/11]^(7/6)] / (5/11)^(4/6) =
11 * (5/11)^(3/6) =
11 * (5/11)^(1/2) =
11*√5 / √11 =
√11* √5 =
√55
