Given that, $ a_{1}, a_{2}, \dots , a_{52} $ be a positive integers, such that $ a_{1} < a_{2} < \dots < a_{52}.$
Now, $ \dfrac{a_{1}+a_{2}+ \dots +a_{52}}{52} = \dfrac{a_{2}+a_{3}+ \dots + a_{52}}{51} – 1 $
$ \Rightarrow \dfrac{a_{1}+a_{2}+ \dots +a_{52}}{52} = \dfrac{a_{2}+a_{3}+ \dots + a_{52}-51}{51} $
$ \Rightarrow 51( a_{1}+a_{2}+ \dots + a_{52}) = 52 (a_{2}+a_{3}+ \dots + a_{52} – 51) $
$ \Rightarrow 51( a_{1}+a_{2}+ \dots + a_{52}) = 52 (a_{2}+a_{3}+ \dots + a_{52}) -(51 \times 52)$
$ \Rightarrow 51a_{1} – (a_{2}+a_{3}+ \dots + a_{52}) = -51 \times 52 $
$ \Rightarrow a_{2}+a_{3}+ \dots + a_{52} = 51a_{1}+ (51 \times 52)$
$ \Rightarrow a_{2}+a_{3}+ \dots + 100 = 51(a_{1}+52) \quad \longrightarrow (1)\quad [ \because a_{52} = 100] $
For largest possible value of $a_{1}:$
$a_{2} = 50, a_{3} = 51, a_{4} = 52, \dots $
From the equation $(1),$ we get
$ 50+51+52+ \dots + 100 = 51 (a_{1}+52) $
$ \Rightarrow 51(a_{1}+52) = \frac{51}{2} (50 + 100)$
$ \Rightarrow a_{1} + 52 = \frac{150}{2} $
$ \Rightarrow a_{1} + 52 = 75 $
$ \Rightarrow a_{1} = 75 – 52 $
$ \Rightarrow \boxed{a_{1} = 23} $
$\therefore$ The largest possible value of $a_{1}$ is $23.$
Correct Answer $: \text{B}$