CAT 2018

Slot 2

QA

Question & Solution

Let 'x' be the average of all 52 positive integers $a_{1}$ , $a_{2}$ ... $a_{52}$ .

$a_{1}+a_{2}+a_{3}+...+a_{52}$ = $52x$ ......................... (1)

Therefore, average of $a_{2}$, $a_{3}$, .... $a_{52}$ = $x+1$

$a_{2}+a_{3}+a_{4}+...+a_{52}$ = $51(x+1)$  ........................ (2)

From equation (1) and (2), we can say that

$a_{1}+51(x+1)$ = 52x

$a_{1}$ = x - 51. 

We have to find out the largest possible value of $a_{1}$. $a_{1}$ will be maximum when 'x' is maximum. 

(x+1) is the average of terms $a_{2}$, $a_{3}$, ....$a_{52}$. We know that $a_{2} \lt a_{3} \lt ... \lt a_{52} $ and  $a_{52}$ = 100

Therefore, (x+1) will be maximum when each term is maximum possible. If $a_{52}$ = 100, then $a_{52}$ = 99, $a_{50}$ = 98 ends so on.

$a_{2}$ = 100 + (51-1)*(-1) = 50.

Hence,  $a_{2}+a_{3}+a_{4}+...+a_{52}$ = 50 + 51 + ....... + 99 + 100 = 51 (x+1)

$\Rightarrow$ $\dfrac{51*(50+100)}{2} = 51(x+1)$

$\Rightarrow$ $x = 74$

Therefore, the largest possible value of $a_{1}$ = x - 51 = 74 - 51 = 23