# precal

what is the 7th geometric sequence when a10 is 9 and a13 is -72

1. 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

posted by Damon
2. 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

posted by Reiny

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