# Find the four numbers forming a G.P. in which the $3^{rd}$ term is greater than the $1^{st}$ term by $9$ and the $2^{nd}$ term is greater than the $4^{th}$ term by $18$.

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

Toolbox:
• $a^2-b^2=(a-b)(a+b)$
• $a^3-b^3=(a-b)(a^2+ab+b^2)$
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 $3,-6,12,-24$