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anonymous
 one year ago
Find the 6th term of a geometric sequence t3 = 444 and t7 = 7104.
7104/444=16
anonymous
 one year ago
Find the 6th term of a geometric sequence t3 = 444 and t7 = 7104. 7104/444=16

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here we have to find the constant of our sequence

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1since your problem asks for 6 terms, and we already have two terms, then we have to add other four terms, between those two terms, namely 444 and 7104. In other words we have to find four numbers, such that: \[\Large 444,\;{a_1},\;{a_2},\;{a_3},\;{a_4},\;7104\] are a geometric sequence

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1sorry since t2= 444 and t7= 7104, we have to find only 3 terms not four, so we have this: \[\Large 444,\;{t_4},\;{t_5},\;{t_6},\;7104\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1next: using the general formula for the nth term of a geometric sequence whose constant is q, we can write this: \[\Large 7104 = 444 \cdot {q^4}\] what is q?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large \sqrt[4]{{16}} = ...?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1correct! so we have: \[\Large {t_4} = 2 \cdot 444 = 888\] and so on...
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