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Q.
In a G.P., if third term is greater than the first term by 9 , and the second term is greater than the fourth term by 18 , then the first four terms of this G.P. are
Sequences and Series
Solution:
Let the G.P. is $a, a r, a r^2, a r^3, \ldots$
Given, third term $=$ First term $+9$
i.e., $ T_3 =a+9$
$ \Rightarrow a r^2 =a+9 $
$ \Rightarrow a r^2-a =9....$(i)
Again, second term $=$ Fourth term $+18$
i.e.,$ T_2=T_4+18$
$\Rightarrow a r=a r^3+18$
$\Rightarrow a r-a r^3=18 ....$(ii)
Dividing Eq. (i) by Eq. (ii), we get
$ \frac{a r^2-a}{a r-a r^3}=\frac{9}{18} $
$ \Rightarrow \frac{a\left(r^2-1\right)}{a r\left(1-r^2\right)}=\frac{1}{2}$
$\Rightarrow \frac{-1}{r} \frac{\left(1-r^2\right)}{\left(1-r^2\right)}=\frac{1}{2}$
$ \Rightarrow -\frac{1}{r}=\frac{1}{2} \Rightarrow r=-2$
Putting $r=-2$ in Eq. (ii), we get
$ a(-2)-a(-2)^3 =18$
$\Rightarrow -2 a+8 a =18 $
$\Rightarrow 6 a =18$
$\Rightarrow a =3$
G.P. is $3,3(-2), 3(-2)^2, 3(-2)^3, \ldots \ldots$
i.e., $ 3,-6,12,-24, \ldots \ldots$