Explanation

(b)
$\because 13,\ a_{1},\ a_{2},\ldots,a_{20},\ 67$ are in AP
$\therefore\ a_{1} + \ a_{2} + \ a_{3} + ... + \ a_{20} = 20\left( \frac{13 + 67}{2} \right) = 800$
Also, AM > GM
$\Rightarrow \ \ \frac{a_{1} + \ a_{2} + \ a_{3} + ... + \ a_{20}}{20} \geq \left( a_{1}\ a_{2}\ a_{3}...a_{20} \right)^{1/20}$
$\Rightarrow 40 \geq \left( a_{1} \bullet \ a_{2} \bullet \ a_{3}...,\ a_{20} \right)^{1/20}$
Hence, maximum value of $a_{1} \bullet \ a_{2} \bullet \ a_{3}...,\ a_{20}$ is $(40)^{20}$