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 a2,a3,....a51 The only way to do this is to assume that a2,a3...a52 are in an AP with a common difference of 1. Let the average of a2,a3....a52 i.e. a27 be 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 a52=a27+25 and asz =100 ⇒A=100−25=75