A geometric sequence (or two ?)

The three terms of this geometric sequence are $a , a r , a r^2$ , therefore,

$104 = a (1 + r + r^2)$ and $13824 = a^3 r^3$

The second equation gives $a r = \sqrt[3]{13824} = 24$ , hence,

$104 = \dfrac{24}{r} (1 + r + r^2)$ , then

$26 r = 6 (1 + r + r^2) \Rightarrow 13 r = 3 (1 + r + r^2) \Rightarrow 3 r^2 - 10 r + 3 = 0 \Rightarrow (3 r - 1)(r - 3 ) = 0 \Rightarrow r = \frac{1}{3} \text{ or } r = 3$

If $r = \dfrac{1}{3}$ then the smallest term is $a r^2 = (a r ) r = 24 (\frac{1}{3}) = 8$ , and if $r = 3$ then the smallest term is $a = \dfrac{ar}{r} = \dfrac{24}{3} = 8$

So in either case the answer is $\boxed {8}$ .

Let $r$ be the common ratio of the geometric progression and $a$ , $b$ and $c$ the three first terms. Then $a=\dfrac{b}{r}$ and $c=br$ , thus we have

$abc=13824\Rightarrow \frac{b}{r}b\left( br \right)=13824\Leftrightarrow {{b}^{3}}=13824\Leftrightarrow b=24$ Hence $ac=\dfrac{1384}{24}=576 \ \ \ \ \ (1)$ And $a+c=104-24=80 \ \ \ \ \ (2)$ $(1)$ and $(2)$ imply that $a$ and $c$ are the roots of the quadratic equation ${{x}^{2}}-80x+576=0$ i.e. $a=8 \ \ \ \ and \ \ \ \ c=72$ or vice versa.

Hence, the smallest term is $\boxed{8}$ .