Answer:
What you're solving forThe \(16\mathrm{th}\) term of the geometric sequence.What’s given in the problemThe first term of the sequence is \(\frac{1}{2}\)The common ratio of the sequence is \(\sqrt[3]{5}\)Helpful informationThe \(n\mathrm{th}\) term of a geometric sequence with the first term \(a_{1}\) and the common ratio \(r\) is given by \(a_{n}=a_{1}r^{n-1}\)Step 1Substitute \(a_{1}=\frac{1}{2}\), \(r=\sqrt[3]{5}\) and \(n=16\) into the formula for the \(n\mathrm{th}\) term of a geometric sequence.\(a_{n}=a_{1}r^{n-1}\)\(a_{16}=\frac{1}{2}(\sqrt[3]{5})^{16-1}\)Step 2 Simplify the expression obtained on the right hand side.
Subtract the numbers in the exponent.\(a_{16}=\frac{1}{2}(\sqrt[3]{5})^{15}\)
Reduce the index of the radical and the exponent by \(3\).\(a_{16}=\frac{1}{2}\cdot 5^{5}\)Calculate the product.\(a_{16}=\frac{5^{5}}{2}\)
Evaluate the power.\(a_{16}=\frac{3125}{2}\)
Solutions The \(16\mathrm{th}\)
term of the sequence is \(\frac{3125}{2}\).