Step 1

The given sequence is exponential, it has n in the exponent. So this is a geometric sequence.

Common ratio of the sequence is:

\(\frac{second\ term}{first\ term}=\frac{\left(\frac{5}{8}\right)^{2}}{\left(\frac{5}{8}\right)^{1}}=\frac{5}{8}\)

Step 2

We will use the \(\sum\) formula of the geometric series.

\(S_{n}=\frac{a^{1}(1\ -\ r^{n})}{1\ -\ r}\)

\(S_{50}=\frac{\frac{5}{8}\left(1\ -\ \left(\frac{5}{8}\right)^{50}\right)}{1\ -\ \frac{5}{8}}\)

\(S_{50}=\ \frac{\frac{5}{8}\ (0.999999999938)}{\frac{3}{8}},\ S_{50}=\ \frac{5(0.999999999938)}{3}\)

\(S_{50}=1.667\)

Answer: Geometric, \(\text{common ratio}=\frac{5}{8},\ \sum=1.667\)

The given sequence is exponential, it has n in the exponent. So this is a geometric sequence.

Common ratio of the sequence is:

\(\frac{second\ term}{first\ term}=\frac{\left(\frac{5}{8}\right)^{2}}{\left(\frac{5}{8}\right)^{1}}=\frac{5}{8}\)

Step 2

We will use the \(\sum\) formula of the geometric series.

\(S_{n}=\frac{a^{1}(1\ -\ r^{n})}{1\ -\ r}\)

\(S_{50}=\frac{\frac{5}{8}\left(1\ -\ \left(\frac{5}{8}\right)^{50}\right)}{1\ -\ \frac{5}{8}}\)

\(S_{50}=\ \frac{\frac{5}{8}\ (0.999999999938)}{\frac{3}{8}},\ S_{50}=\ \frac{5(0.999999999938)}{3}\)

\(S_{50}=1.667\)

Answer: Geometric, \(\text{common ratio}=\frac{5}{8},\ \sum=1.667\)