(a) The third term of the arithmetic progression is \(20+2d\), and the third term of the geometric progression is \(20r^2\).

So

\( 20r^2=20+2d+5 \).

\( 20r^2=25+2d \). (1)

The fifth term of the arithmetic progression is \(20+4d\), and the fifth term of the geometric progression is \(20r^4\).

So

\( 20r^4=20+4d+30 \).

\( 20r^4=50+4d \). (2)

From (1), \( d=10r^2-\frac{25}{2} \).

From (2), \( d=5r^4-\frac{25}{2} \).

Equating these:

\( 10r^2-\frac{25}{2}=5r^4-\frac{25}{2} \).

\( 10r^2=5r^4 \).

Since \(r>0\), \( r^2=2 \), so \( r=\sqrt2 \).

Then

\( d=10(2)-\frac{25}{2}=\frac{15}{2} \).

Answer: \( r=\sqrt2,\ d=\frac{15}{2} \).

(b) The ninth term of the geometric progression is

\( 20r^8=20(\sqrt2)^8=20(16)=320 \).

The ninth term of the arithmetic progression is

\( 20+8d=20+8\left(\frac{15}{2}\right)=80 \).

Since \(320=4(80)\), the ninth term of the geometric progression is \(4\) times the ninth term of the arithmetic progression.