Write a formula for the general or term for each geometric sequence.
$12,-6,3,-\frac{3}{2},\dots$

• A. $a_n=12{(-2)}^{n-1}$
• B. $a_n=12{(-\frac{1}{2})}^{n-1}$
• C. $a_n=12{\left(\frac{1}{2}\right)}^{n-1}$
• D. $a_n={(-\frac{1}{2})}^{n-1}$

Answer: (B)

Verified step by step guidance

1. Identify the first term of the geometric sequence, which is the initial value of the sequence. In this case, the first term \(a_1\) is 12.
2. Determine the common ratio \(r\) by dividing the second term by the first term. Calculate \(r = \frac{a_2}{a_1} = \frac{-6}{12}\).
3. Simplify the common ratio to get \(r = -\frac{1}{2}\). This ratio tells us how each term relates to the previous term by multiplication.
4. Write the general formula for the \(n^{\operatorname{th}}\) term of a geometric sequence, which is \(a_n = a_1 \times r^{n-1}\).
5. Substitute the values of \(a_1\) and \(r\) into the formula to get \(a_n = 12 \left(-\frac{1}{2}\right)^{n-1}\), which represents the \(n^{\operatorname{th}}\) term of the sequence.