Let the first term be \(a\) and the common ratio be \(r\).

The sum of the first two terms is given by:

\(a(1 + r) = 15\)

The sum to infinity is given by:

\(\frac{a}{1 - r} = \frac{125}{7}\)

From the first equation, we have:

\(a = \frac{15}{1 + r}\)

Substitute \(a\) in the second equation:

\(\frac{15}{1 + r} \cdot \frac{1}{1 - r} = \frac{125}{7}\)

\(\frac{15}{(1 + r)(1 - r)} = \frac{125}{7}\)

\(\frac{15}{1 - r^2} = \frac{125}{7}\)

\(15 \cdot 7 = 125(1 - r^2)\)

\(105 = 125 - 125r^2\)

\(125r^2 = 20\)

\(r^2 = \frac{4}{25}\)

Since \(r\) is negative, \(r = -\frac{2}{5}\).

Substitute \(r\) back to find \(a\):

\(a = \frac{15}{1 - \frac{2}{5}} = \frac{15}{\frac{3}{5}} = 25\)

The third term is \(ar^2\):

\(ar^2 = 25 \cdot \frac{4}{25} = 4\)