To find a formula for the given sequence \(-99, \frac{198}{5}, -\frac{396}{25}\), we will express the \(n\)-th term of the sequence \(a_n\) in the form \(a_n = a \cdot r^{n-1}\), where \(a\) is the first term and \(r\) is the common ratio.
-
First Term \(a\):
- The first term \(a_1\) is \(-99\). Therefore, \(a = -99\).
-
Finding the Common Ratio \(r\):
- To find \(r\), we divide the second term by the first term: \[ r = \frac{a_2}{a_1} = \frac{\frac{198}{5}}{-99} \] To simplify this, we can multiply the numerator and denominator: \[ r = \frac{198}{5} \cdot \frac{-1}{99} = \frac{198 \cdot -1}{99 \cdot 5} = \frac{-198}{495} \] Simplifying \(\frac{-198}{495}\):
- \(198\) and \(495\) can both be divided by \(99\): \[ r = \frac{-2}{5} \]
-
Formulating the nth Term:
- We have \(a = -99\) and \(r = -\frac{2}{5}\). Thus, the formula for \(a_n\) becomes: \[ a_n = -99 \cdot \left(-\frac{2}{5}\right)^{n-1} \]
-
Final Expression:
- Therefore, the equation that describes the sequence is: \[ a_n = -99 \left(-\frac{2}{5}\right)^{n-1} \]
Final answer: \[ a_n = -99 \left(-\frac{2}{5}\right)^{n-1} \]