Answer: first term \(10\), common ratio \(\frac12\).

Answer: first term \(10\), common ratio \(\frac12\).

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

The sum of the first three terms is

\(a+ar+ar^2=17.5.\)

The sum to infinity is

\(\frac{a}{1-r}=20.\)

Therefore

\(a=20(1-r).\)

Substitute this into the first-three-terms equation:

\(20(1-r)(1+r+r^2)=17.5.\)

Since

\((1-r)(1+r+r^2)=1-r^3,\)

we get

\(20(1-r^3)=17.5.\)

So

\(1-r^3=0.875, \qquad r^3=0.125=\frac18.\)

Hence

\(r=\frac12.\)

Then

\(a=20\left(1-\frac12\right)=10.\)