CAT 2018 Question Paper Shift 2 Solved Paper

We want to maximize the value of a 1 , subject to the condition that al is the least of the 52 numbers and that the average of 51 numbers (excluding a 1 ) is 1 less than the average of all the 52 numbers. Since as2 is 100 and all the numbers are positive integers, maximizing al entails maximizing ${\mathit{a}}_2,{\mathit{a}}_3,....{\mathit{a}}_{51}$
The only way to do this is to assume that ${\mathit{a}}_2,{\mathit{a}}_3...{\mathit{a}}_{52}$ are in an $\mathit{A}\mathit{P}$ with a common difference of $1.$
Let the average of ${\mathit{a}}_2,{\mathit{a}}_3....{\mathit{a}}_{52}$ i.e. ${\mathit{a}}_{27}$ be $\mathit{A}$.
(Note: The average of an odd number of terms in an Arithmetic Progression is equal to the value of the middle-most term)
Since ${\mathit{a}}_{52}={\mathit{a}}_{27}+25$ and asz $=100$
$\Rightarrow \mathit{A}=100-25=75$
Given Hence ${\mathit{a}}_1=3848-3825=23$