Let the first term be \(a\) and the common ratio be \(r\).
The second term is given by \(ar = 54\).
The sum to infinity is given by \(\frac{a}{1-r} = 243\).
From \(ar = 54\), we have \(a = \frac{54}{r}\).
Substitute \(a\) in the sum to infinity equation:
\(\frac{54}{r(1-r)} = 243\)
\(54 = 243r(1-r)\)
\(243r^2 - 243r + 54 = 0\)
Divide through by 9:
\(27r^2 - 27r + 6 = 0\)
Solving this quadratic equation using the quadratic formula:
\(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(r = \frac{27 \pm \sqrt{27^2 - 4 \times 27 \times 6}}{2 \times 27}\)
\(r = \frac{27 \pm \sqrt{729 - 648}}{54}\)
\(r = \frac{27 \pm \sqrt{81}}{54}\)
\(r = \frac{27 \pm 9}{54}\)
\(r = \frac{36}{54} = \frac{2}{3}\) or \(r = \frac{18}{54} = \frac{1}{3}\)
Since \(r > \frac{1}{2}\), we choose \(r = \frac{2}{3}\).
Substitute \(r = \frac{2}{3}\) back to find \(a\):
\(ar = 54\)
\(a \times \frac{2}{3} = 54\)
\(a = 54 \times \frac{3}{2} = 81\)
The tenth term is given by \(ar^9\):
\(ar^9 = 81 \left( \frac{2}{3} \right)^9\)
\(ar^9 = 81 \times \frac{512}{19683}\)
\(ar^9 = \frac{81 \times 512}{19683}\)
\(ar^9 = \frac{512}{243}\)
