$\begin{array}{1 1}3,-9,27,-81 \\ 2,-6,18,-54 \\ 3,6,12,24 \\ 3,-6,12,-24 \end{array} $

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- $a^2-b^2=(a-b)(a+b)$
- $a^3-b^3=(a-b)(a^2+ab+b^2)$

Let the four terms in G,P, be $a,ar,ar^2\:\:and\:\:ar^3$

It is given that $3^{rd}$ term is greater than the $1^{st}$ term by $9$.

and

the $2^{nd}$ term is greater than the $4^{th}$ term by $18$.

$i.e., t_1+9=t_3$ and $t_4+18=t_2$

$\Rightarrow\:a+9=ar^2$.....(i) and

$ar^3+18=ar$.........(ii)

$\Rightarrow\:a(r^2-1)=9\:\:\:and\:\:\:a(r-r^3)=18$

Dividing the two equations

$\Rightarrow\:\large\frac{ar(1-r^2)}{a(r^2-1)}=\frac{18}{9}$

Taking - sign common from numerator

$\Rightarrow\:-\large\frac{ar(r^2-1)}{a(r^2-1)}=\frac{18}{9}$

$\Rightarrow\:r=-2$

Substituting the value of $r$ in (i) we get

$\Rightarrow\:a=\large\frac{9}{(-2)^2-1}=\frac{9}{3}$$=3$

Substituting the values of $a\:\:and\:\:r$ we get

$ar=3\times (-2)=-6,\:\:ar^2=3\times (-2)^2=12\:\;and\:\:ar^3=3\times(-2)^3=-24$

$\therefore$ The four numbers in G.P are $3,-6,12,-24$

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