What is a formula for the nth term of the given sequence? \[ \begin{array}{ll}4,2,1 \ldots\end{array} \] Answer \( \begin{array}{ll}a_{n}=8\left(\frac{1}{2}\right)^{-n} & a_{n}=4\left(\frac{1}{2}\right)^{n-1} \\ a_{n}=4(2)^{n} & a_{n}=8(2)^{n-1}\end{array} \)

The sequence you're looking at is a geometric series where each term is halved from the previous one. The formula for the nth term can actually be derived from the first term and the common ratio, which here is \( \frac{1}{2} \). So, the correct formula for the nth term is \( a_n = 4\left(\frac{1}{2}\right)^{n-1} \). In terms of real-world applications, geometric sequences like this one can be useful in predicting situations involving exponential decay, such as in population dynamics or radioactive decay, where quantities halve over equal time intervals. Understanding these concepts can help in fields ranging from environmental science to finance!