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Let $a_{1},a_{2},\dots , a_{52}$ be a positive integers such that $a_{1}<a_{2}<\dots < a_{52}$. Suppose, their arithmetic mean is one less than the arithmetic mean of  $a_{2},a_{3},\dots , a_{52}$. If $a_{52}=100$  , then the largest possible value of $a_{1}$ is

  1. $20$
  2. $23$
  3. $48$
  4. $45$
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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 $: \text{B}$
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