The common ratio of a GP is 3 and the but term is 486 I

SOLUTION

r = 3
Last term $=436=a{r}^{n-1}=a{3}^{n-1}$
$S_n=723=\frac{a(1-r^n)}{1-r}=\frac{a(r^n-1)}{r-1}$
$\frac{a}{2}(3^n-1)$
$\frac{723}{436}=\frac{2\text{/}a{3}^{n-1}}{\text{/}a(3^n-1)}$
From here you can find n and hence you can find a.