precal
posted by gail
what is the 7th geometric sequence when a10 is 9 and a13 is 72

Damon
9 9r 9r^2 9r^3
72 = 9 r^3
r^3 = 72/9
now work back
a9 = 9/r
a8 = 9/r^2
a7 = 9/r^3
interesting
a7 = 9/r^3 = (9)(9/72) = 81/72 
Reiny
or ,
using the standard a to be first term, r the common ratio
ar^12 = 72
ar^9 =9
divide them
r^3 = 8
r = 2 , then from ar^9 = 9
a = 9/(2)^9 = 9/512
term(7) = ar^6
= (9/512)(2^6) = 576/512 = 9/8
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