Geometric Sequence and Recursive Definition

[rought]

Homework Statement

In a Geometric Sequence find t₁₀. t₃ = 4 and t₆= 4/27

Give a recursive definition for the sequence: 1, 4, 13, 40...

Homework Equations

I know that a Geometric sequence is: t_n=t₁(r)^n-1

And that a recursive formula starts off with t_n-1

I'm not sure where to go from here...

 

[Mark44]

A geometric sequence has terms that look like this: {a, ar, ar², ar³, ...}, where each term is r time the previous term. In your first problem, you have t₃ = 4, so t₄ = rt₃ = 4r, t₅ = 4t₄ = ? Just work your way up to the ones you want to find.

In your second problem, you have {1, 4, 13, 40, ...}
The 2nd term is the first plus 3. The 3rd term is 9 plus the 2nd. The 4th term is 27 + the 3rd. Do you see a pattern?

 

[Architeuthis Dux]

To find t10 in a geometric sequence, we can use the formula tn = t1(r)^(n-1). In this case, we are given t3 = 4 and t6 = 4/27. We can use these values to find the common ratio, r.
First, we can rewrite t3 = 4 as t1(r)^2 = 4, and t6 = 4/27 as t1(r)^5 = 4/27.
Dividing the second equation by the first, we get (r)^3 = 1/27.
Taking the cube root of both sides, we get r = 1/3.
Now, we can plug in this value to the original formula, tn = t1(r)^(n-1), to find t10.
t10 = t1(1/3)^9.
Without knowing the value of t1, we cannot find the exact value of t10, but we can express it in terms of t1 as t10 = t1(1/3)^9.

A recursive definition for the sequence 1, 4, 13, 40... can be written as:
t1 = 1
tn = 3tn-1 + 1, for n > 1
This means that the first term, t1, is equal to 1, and each subsequent term is found by multiplying the previous term by 3 and adding 1. This recursive definition can also be written as tn = 3^(n-1) + 1.