Question

question
what is a formula for the nth term of the given sequence?
75, 45, 27...
answer
$a_n = 125(\frac{3}{5})^{-n}$ $a_n = 125(\frac{3}{5})^{n - 1}$
$a_n = 125(\frac{5}{3})^{n - 1}$ $a_n = 125(\frac{5}{3})^{-n}$

Explanation:

Step1: Identify the type of sequence

The given sequence \(75,45,27,\cdots\) is a geometric sequence since \(\frac{45}{75}=\frac{3}{5}\) and \(\frac{27}{45}=\frac{3}{5}\), with a common - ratio \(r = \frac{3}{5}\).

Step2: Recall the formula for the nth term of a geometric sequence

The formula for the nth term of a geometric sequence is \(a_{n}=a_{1}r^{n - 1}\), where \(a_{1}\) is the first - term and \(r\) is the common ratio. Here, \(a_{1}=75\).

Step3: Express \(a_{1}\) in a different form

We know that \(75 = 125\times\frac{3}{5}\). So, \(a_{n}=125\times\frac{3}{5}\times(\frac{3}{5})^{n - 1}\).

Step4: Simplify the formula

Using the exponent rule \(a^{m}\times a^{n}=a^{m + n}\), we have \(a_{n}=125\times(\frac{3}{5})^{n - 1+1}=125(\frac{3}{5})^{n - 1}\).

Answer:

\(a_{n}=125(\frac{3}{5})^{n - 1}\)