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

1. anonymous

@Michele_Laino

2. anonymous

16/2=8

3. Michele_Laino

here we have to find the constant of our sequence

4. anonymous

okay

5. 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

6. anonymous

okay

7. Michele_Laino

sorry since t2= 444 and t7= 7104, we have to find only 3 terms not four, so we have this: $\Large 444,\;{t_4},\;{t_5},\;{t_6},\;7104$

8. 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?

9. Michele_Laino

oops..t3=444

10. anonymous

q4=16

11. Michele_Laino

so, q=...?

12. anonymous

4

13. Michele_Laino

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

14. anonymous

2

15. Michele_Laino

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

16. anonymous

3552