The sum of first four terms of a G.P. is 160 and the common ratio is 3 , then the $4^{\text {th }}$ term is
The sum of first four terms of a G.P. is 160 and the common ratio is 3 , then the $4^{\text {th }}$
term is
118
100
108
102
Solution
$a, a r, a r^{2}, a r^{3}$
$2=3$
$\frac{a\left(\varepsilon^{4}-1\right)}{(\varepsilon-1)}=160$
Solving these we get $a=4$
$a \varepsilon^{3}=t_{4}=108$