Answer: \(r=\frac45\), first term \(5\); or \(r=-\frac45\), first term \(45\).

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

The sum of the first two terms is

\(a+ar=a(1+r).\)

Since this is \(9\),

\(a(1+r)=9. \tag{1}\)

The sum to infinity of a geometric progression is

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

where \(|r|\lt1\). It is given that this sum is \(25\), so

\(\dfrac{a}{1-r}=25.\)

Hence

\(a=25(1-r). \tag{2}\)

Substitute (2) into (1):

\(25(1-r)(1+r)=9.\)

So

\(25(1-r^2)=9.\)

Therefore

\(1-r^2=\dfrac9{25},\qquad r^2=\dfrac{16}{25}.\)

Thus

\(r=\dfrac45\quad\text{or}\quad r=-\dfrac45.\)

Both satisfy \(|r|\lt1\), so both are possible.

If \(r=\dfrac45\), then

\(a=25\left(1-\dfrac45\right)=5.\)

If \(r=-\dfrac45\), then

\(a=25\left(1+\dfrac45\right)=45.\)

Therefore the possibilities are \(r=\dfrac45\) with first term \(5\), or \(r=-\dfrac45\) with first term \(45\).