Question

For each sequence, find a formula for the general term, \( a_{n} \). \[ \begin{array}{l} \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots a_{n}=\square \\ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots \quad a_{n}=\square \end{array} \] \[ 4,6,8,10, \ldots \quad a_{n}= \] \( \square \)

          For each sequence, find a formula for the general term, \( a_{n} \).
\[
\begin{array}{l}
\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots a_{n}=\square \\
\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots \quad a_{n}=\square
\end{array}
\]
\[
4,6,8,10, \ldots \quad a_{n}=
\]
\( \square \)

For each sequence, find a formula for the general term, an.

(1)/(2), (1)/(4), (1)/(8), (1)/(16), … an=□
(1)/(2), (2)/(3), (3)/(4), (4)/(5), … an=□

4,6,8,10, … an=

□

Added by Mark S.

Algebra and Trigonometry

Judith A. Beecher, Judith A. Penna,… 4th Edition

Chapter 11

Instant Answer

Solved by Expert Gregory Higby

Step 1

The first term \( a_1 \) is \( \frac{1}{2} \), and the common ratio \( r \) is \( \frac{1}{2} \). Step 2: The formula for the \( n \)-th term of a geometric sequence is \( a_n = a_1 \cdot r^{(n-1)} \). Step 3: Substitute \( a_1 = \frac{1}{2} \) and \( r = Show more…

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For each sequence, find a formula for the general term, \( a_{n} \). \[ \begin{array}{l} \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \ldots a_{n}=\square \\ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots \quad a_{n}=\square \end{array} \] \[ 4,6,8,10, \ldots \quad a_{n}= \] \( \square \)