Question

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

7104/444=16

Answer 1

@Michele_Laino

Answer 2

16/2=8

Answer 3

here we have to find the constant of our sequence

Answer 4

okay

Answer 5

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

Answer 6

okay

Answer 7

sorry since t2= 444 and t7= 7104, we have to find only 3 terms not four, so we have this:

\[\Large 444,\;{t_4},\;{t_5},\;{t_6},\;7104\]

Answer 8

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?

Answer 9

oops..t3=444

Answer 10

q4=16

Answer 11

so, q=...?

Answer 12

4

Answer 13

\[\Large \sqrt[4]{{16}} = ...?\]

Answer 14

2

Answer 15

correct! so we have:
\[\Large {t_4} = 2 \cdot 444 = 888\]
and so on...

Answer 16

3552