Formulas for Geometric Sequences: Recursive and Explicit
Core Concepts
Master the recursive and explicit formulas for geometric sequences. Learn to apply these formulas to find specific terms, write rules for sequences, and solve for missing values like the first term or common ratio.
A geometric sequence has \(a_3 = 20\) and \(a_6 = 160\). Find the common ratio, \(r\).
Use \(a_m = a_k \cdot r^{m-k}\). We have \(160 = 20 \cdot r^{6-3}\), which simplifies to \(8 = r^3\). Therefore, \(r = 2\).
Given that a geometric sequence has \(r=2\) and \(a_3=20\), find the first term, \(a_1\).
Using \(a_n = a_1 \cdot r^{n-1}\), we get \(20 = a_1 \cdot 2^{3-1}\), which is \(20 = a_1 \cdot 4\). So, \(a_1 = 5\).
In the geometric sequence with \(a_1 = 3\) and \(r = 4\), which term number \(n\) has the value \(3072\)?
Set \(a_n = 3072\): \(3072 = 3 \cdot 4^{n-1}\). Divide by \(3\) to get \(1024 = 4^{n-1}\). Since \(4^5 = 1024\), we have \(n-1 = 5\), so \(n = 6\).
What is the geometric mean of two positive numbers, \(x\) and \(y\)?
The geometric mean is \(\sqrt{xy}\). It is the term that would be located exactly in the middle of \(x\) and \(y\) in a geometric sequence.
Find the positive missing term in the geometric sequence: \(4, ?, 36\).
The missing term is the geometric mean of \(4\) and \(36\): \(\sqrt{4 \cdot 36} = \sqrt{144} = 12\).
How does the explicit formula for a geometric sequence relate to an exponential function \(y = ab^x\)?
The formula \(a_n = a_1 \cdot r^{n-1}\) is a form of an exponential function where the common ratio \(r\) is the base, and the domain is restricted to the positive integers.
What is the general form of a **recursive formula** for a geometric sequence?
\(a_n = a_{n-1} \cdot r\). To be complete, the formula must also state the value of the first term, \(a_1\).
Write the recursive formula for the sequence \(2, 10, 50, 250, ...\).
The first term is \(a_1 = 2\) and the common ratio is \(r = 5\). The recursive formula is \(a_1 = 2\) and \(a_n = a_{n-1} \cdot 5\).
What is the general form of the **explicit formula** for a geometric sequence?
\(a_n = a_1 \cdot r^{n-1}\), where \(a_1\) is the first term and \(r\) is the common ratio.
In the explicit formula \(a_n = a_1 \cdot r^{n-1}\), why is the exponent \(n-1\) instead of \(n\)?
Because the first term (\(a_1\)) is the starting point. To get to the \(n\)-th term, you only need to perform the multiplication by \(r\) a total of \(n-1\) times.
Write the explicit formula for the geometric sequence \(4, 8, 16, 32, ...\).
The first term is \(a_1 = 4\) and the common ratio is \(r = 2\). The explicit formula is \(a_n = 4 \cdot 2^{n-1}\).
