Answer: \(r=4\); first term \(\frac1{16}\); no sum to infinity because \(r\gt 1\).

Let the first term be \(a\) and the common ratio be \(r\). Since \(r\ne1\),

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

Given \(S_4=17S_2\),

\(\frac{a(r^4-1)}{r-1}=17\frac{a(r^2-1)}{r-1}.\)

Canceling the common factors gives

\(r^4-1=17(r^2-1).\)

Using \(r^4-1=(r^2-1)(r^2+1)\), and since \(r\ne1\),

\(r^2+1=17.\)

So

\(r^2=16.\)

The common ratio is positive, hence

\(r=4.\)

The \(6\)th term is \(ar^5\). Since it is \(64\),

\(a(4^5)=64.\)

Therefore

\(a=\frac{64}{1024}=\frac1{16}.\)

A geometric progression has a sum to infinity only when \(|r|\lt 1\). Here \(r=4\gt 1\), so this progression does not have a sum to infinity.