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

The third term is \(ar^2\) and the sixth term is \(ar^5\).

Given that \(ar^2 = 8 \times ar^5\), we have:

\(ar^2 = 8ar^5\)

\(r^2 = 8r^5\)

\(r^3 = \frac{1}{8}\)

\(r = \frac{1}{2}\)

The sum of the first six terms is given by:

\(S_6 = \frac{a(1-r^6)}{1-r} = 31.5\)

\(\frac{a(1-(\frac{1}{2})^6)}{\frac{1}{2}} = 31.5\)

\(\frac{a(1-\frac{1}{64})}{\frac{1}{2}} = 31.5\)

\(\frac{a(\frac{63}{64})}{\frac{1}{2}} = 31.5\)

\(\frac{63a}{64} \times 2 = 31.5\)

\(\frac{63a}{32} = 31.5\)

\(63a = 31.5 \times 32\)

\(63a = 1008\)

\(a = 16\)

The sum to infinity is given by:

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

\(S_\infty = \frac{16}{1-\frac{1}{2}}\)

\(S_\infty = \frac{16}{\frac{1}{2}}\)

\(S_\infty = 32\)