anonymous
  • anonymous
Find the 6th term of a geometric sequence t3 = 444 and t7 = 7104. 7104/444=16
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@Michele_Laino
anonymous
  • anonymous
16/2=8
Michele_Laino
  • Michele_Laino
here we have to find the constant of our sequence

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anonymous
  • anonymous
okay
Michele_Laino
  • Michele_Laino
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
anonymous
  • anonymous
okay
Michele_Laino
  • Michele_Laino
sorry since t2= 444 and t7= 7104, we have to find only 3 terms not four, so we have this: \[\Large 444,\;{t_4},\;{t_5},\;{t_6},\;7104\]
Michele_Laino
  • Michele_Laino
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?
Michele_Laino
  • Michele_Laino
oops..t3=444
anonymous
  • anonymous
q4=16
Michele_Laino
  • Michele_Laino
so, q=...?
anonymous
  • anonymous
4
Michele_Laino
  • Michele_Laino
\[\Large \sqrt[4]{{16}} = ...?\]
anonymous
  • anonymous
2
Michele_Laino
  • Michele_Laino
correct! so we have: \[\Large {t_4} = 2 \cdot 444 = 888\] and so on...
anonymous
  • anonymous
3552

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