Problem
A sequence \(\{a_n\}\) is geometric with first term \(a_1=81\) and common ratio \(r=\frac{1}{3}\). Determine the recursive formula for geometric sequence and compute the 6th term \(a_6\).
Step 1: Recall what “geometric” means
A geometric sequence has a constant ratio between consecutive terms: \[ \frac{a_n}{a_{n-1}} = r \quad (n \ge 2). \] Here, the common ratio is \(r=\frac{1}{3}\), so each term is one third of the previous term.
Step 2: Write the recursive formula
A recursive definition needs:
- A starting value (base case).
- A recurrence rule describing \(a_n\) using earlier term(s).
The standard recursive form for a geometric sequence is: \[ a_1 = \text{(first term)}, \qquad a_n = r\,a_{n-1}\ \text{for}\ n \ge 2. \]
Substituting \(a_1=81\) and \(r=\frac{1}{3}\) gives the recursive formula: \[ a_1 = 81, \qquad a_n = \frac{1}{3}\,a_{n-1}\ \text{for}\ n \ge 2. \]
Checkpoint
The recurrence \(a_n=\frac{1}{3}a_{n-1}\) guarantees \(\frac{a_n}{a_{n-1}}=\frac{1}{3}\) for every \(n\ge2\), so the ratio is constant as required for a geometric progression.
Step 3: Compute \(a_6\) using the recursion
Apply the recurrence repeatedly:
\[ a_2=\frac{1}{3}a_1=\frac{1}{3}\cdot 81=27 \] \[ a_3=\frac{1}{3}a_2=\frac{1}{3}\cdot 27=9 \] \[ a_4=\frac{1}{3}a_3=\frac{1}{3}\cdot 9=3 \] \[ a_5=\frac{1}{3}a_4=\frac{1}{3}\cdot 3=1 \] \[ a_6=\frac{1}{3}a_5=\frac{1}{3}\cdot 1=\frac{1}{3}. \]
Optional verification using the explicit formula
The explicit (closed-form) formula for a geometric sequence is: \[ a_n = a_1 \cdot r^{\,n-1}. \] With \(a_1=81\) and \(r=\frac{1}{3}\), \[ a_6 = 81\left(\frac{1}{3}\right)^{5} = 81 \cdot \frac{1}{243} = \frac{1}{3}. \] This matches the recursive computation.
Table of the first terms
| \(n\) | \(a_n\) | Computed from |
|---|---|---|
| 1 | \(81\) | Given |
| 2 | \(27\) | \(a_2=\frac{1}{3}a_1\) |
| 3 | \(9\) | \(a_3=\frac{1}{3}a_2\) |
| 4 | \(3\) | \(a_4=\frac{1}{3}a_3\) |
| 5 | \(1\) | \(a_5=\frac{1}{3}a_4\) |
| 6 | \(\frac{1}{3}\) | \(a_6=\frac{1}{3}a_5\) |
Visualization
Final results
- Recursive formula: \(a_1=81\), \(a_n=\frac{1}{3}a_{n-1}\) for \(n\ge2\).
- 6th term: \(a_6=\frac{1}{3}\).