Let the four numbers in geometric progression (G.P.) be
\[ a,\; ar,\; ar^2,\; ar^3 \]
where \( a \) is the first term and \( r \) is the common ratio.
It is given that the third term is greater than the first term by 9. Hence,
\[ ar^2 - a = 9 \]
\[ a(r^2 - 1) = 9 \quad \text{… (1)} \]
It is also given that the second term is greater than the fourth term by 18. Hence,
\[ ar - ar^3 = 18 \]
\[ ar(1 - r^2) = 18 \]
\[ -ar(r^2 - 1) = 18 \]
\[ ar(r^2 - 1) = -18 \quad \text{… (2)} \]
Dividing equation (2) by equation (1), we get
\[ \frac{ar(r^2 - 1)}{a(r^2 - 1)} = \frac{-18}{9} \]
\[ r = -2. \]
Substituting \( r = -2 \) in equation (1):
\[ a((-2)^2 - 1) = 9 \]
\[ a(4 - 1) = 9 \]
\[ 3a = 9 \]
\[ a = 3. \]
Therefore, the four numbers are
\[ a = 3,\quad ar = -6,\quad ar^2 = 12,\quad ar^3 = -24. \]
Hence, the required geometric progression is
\[ \boxed{3,\; -6,\; 12,\; -24}. \]
If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab) n .