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

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

Mathematics
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16/2=8
here we have to find the constant of our sequence

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okay
since 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
okay
sorry since t2= 444 and t7= 7104, we have to find only 3 terms not four, so we have this: \[\Large 444,\;{t_4},\;{t_5},\;{t_6},\;7104\]
next: using the general formula for the n-th term of a geometric sequence whose constant is q, we can write this: \[\Large 7104 = 444 \cdot {q^4}\] what is q?
oops..t3=444
q4=16
so, q=...?
4
\[\Large \sqrt[4]{{16}} = ...?\]
2
correct! so we have: \[\Large {t_4} = 2 \cdot 444 = 888\] and so on...
3552

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