(c)
Since, $a + ar = a(1 + r) = 12$ …(i)
and $ar^{2} + ar^{3} = ar^{2}(1 + r) = 48$ …(ii)
From Eqs. (i) and (ii),
$r^{2} = 4$
$\Rightarrow \ r = - 2$
(Since, the series is alternately sign, so we take negative values).
On putting the value of $r$ in Eq. (i), we get $a = - 12$