Find the 1st Term and Common Ratio - Algebra-1

1. Fundamental Concepts

First Term (\(a_1\)): The initial term of a geometric sequence, serving as the starting value and foundation for calculating all other terms.
Common Ratio (r): The fixed ratio between consecutive terms in a geometric sequence (from the second term onward), which determines the sequence’s pattern of growth or decay.
Relationship and Role: The first term and common ratio are the two core elements that define a geometric sequence. Given these two values, any term in the sequence can be found using the explicit formula \(a_n = a_1 \cdot r^{n-1}\). Conversely, if certain terms of the sequence are known, we can solve for the first term and common ratio by setting up equations.

Uniqueness of \(a_1\) and r: For a specific geometric sequence, the first term and common ratio are unique (with a special case: the common ratio may have two solutions, positive and negative, when non-adjacent terms are given and the difference between their positions is even).
Using Known Terms: If two terms of the sequence (\(a_m\) and \(a_n\), where \(m \neq n\)) are known, we can set up a system of equations using the explicit formula to solve for \(a_1\) and r.
Constraints on r: The common ratio r cannot be 0 (otherwise, all terms after the first would be 0, making the ratio undefined). The first term \(a_1\) can be 0, but this would make all terms 0 (with \(r = 0\) as well), which is not typically considered a standard geometric sequence.

Question: Given that the 2nd term of a geometric sequence is 6 and the 3rd term is 12, find the first term \(a_1\) and common ratio r.
Solution:
- By definition of the common ratio: \(r = \frac{a_3}{a_2} = \frac{12}{6} = 2\).
- Using the 2nd term: \(a_2 = a_1 \cdot r \implies 6 = a_1 \cdot 2 \implies a_1 = 3\).
Conclusion: \(a_1 = 3\) and \(r = 2\).

Question: In a geometric sequence, the 3rd term is 18, the 5th term is 162, and the common ratio is positive. Find \(a_1\) and r.
Solution:
- Using the explicit formula:\(a_3 = a_1 \cdot r^2 = 18\)\(a_5 = a_1 \cdot r^4 = 162\)
- Divide the second equation by the first: \(\frac{a_5}{a_3} = \frac{r^4}{r^2} = r^2 = \frac{162}{18} = 9\). Since r is positive, \(r = 3\).
- Substitute \(r = 3\) into \(a_3\): \(a_1 \cdot 3^2 = 18 \implies 9a_1 = 18 \implies a_1 = 2\).
Conclusion: \(a_1 = 2\) and \(r = 3\).

Question: In a geometric sequence, the 4th term is -24 and the 6th term is -96. Find \(a_1\) and r.
Solution:
- Using the explicit formula:\(a_4 = a_1 \cdot r^3 = -24\)\(a_6 = a_1 \cdot r^5 = -96\)
- Divide the second equation by the first: \(\frac{a_6}{a_4} = \frac{r^5}{r^3} = r^2 = \frac{-96}{-24} = 4 \implies r = \pm 2\).
- If \(r = 2\): Substitute into \(a_4\): \(a_1 \cdot 2^3 = -24 \implies 8a_1 = -24 \implies a_1 = -3\).
- If \(r = -2\): Substitute into \(a_4\): \(a_1 \cdot (-2)^3 = -24 \implies -8a_1 = -24 \implies a_1 = 3\).
Conclusion: \(a_1 = -3\) and \(r = 2\), or \(a_1 = 3\) and \(r = -2\).

Finding r with adjacent terms: If two adjacent terms (e.g., \(a_n\) and \(a_{n+1}\)) are known, calculate r directly using \(r = \frac{a_{n+1}}{a_n}\), then find \(a_1\) by substituting into one term: \(a_1 = \frac{a_n}{r^{n-1}}\).
Setting up equations for non-adjacent terms: For two known terms \(a_m\) and \(a_n\) (m < n), use \(a_m = a_1 \cdot r^{m-1}\) and \(a_n = a_1 \cdot r^{n-1}\). Divide the two equations to eliminate \(a_1\), resulting in \(r^{n-m} = \frac{a_n}{a_m}\). Solve for r, then substitute back to find \(a_1\).
Handling multiple solutions for r: When solving for r involves squaring (e.g., \(r^2 = k\)), consider both positive and negative values of r. Use additional conditions (e.g., “r is positive”) to filter valid solutions.
Verifying solutions: After finding \(a_1\) and r, substitute them back into the known terms to check if they satisfy the equations, ensuring correctness.