Answer to Question #13034 in Quantitative Methods for Mohit

Question

Let A be a set of distinct positive integers. If the arithmetic mean (average) of the elements of A is 25, what is the maximum possible value of an element in A?

Answer 1

Suppose A consists of n distinct positive integers

a_1,a_2,...,a_n.
Since their mean is 25 we have
that

a_1+a_2+...+a_n=25*n

We can assume that the numbers are
ordered, and so
a_1 < a_2 < ... < a_n.

Hence the maximum
possible value for an will be achieved, when

a_1=1 a_2=2 ...
a_{n-1} = n-1

By formula for the sum of arithmetic progression we have

1+2+...+n-1 = (n-1)(n-2)/2,

Therefore

a_n = 25*n -
(1+2+...+n-1)
= 25 n - (n-1)(n-2)/2
= 25 n - (n^2 + n +2n +2 )
/2
= 25 n - n^2/2 + 1.5n + 1
= - n^2/2 + 26.5n + 1

We
should find maximum of the function
g(n) = - n^2/2 + 26.5n + 1
among all
integer positive numbers.

Let us find critical points of g, ie. solutions
of the equation
g'(n)=0.

We have that
g'(n) = -n + 26.5 = 0

whence
n = 26.5

Hence the maximum of g anomg positive integers
is achieved either at n=26 or at n=27.

Notice that

g(26) = -
26^2/2 + 26.5*26 + 1 = 352
g(27) = - 27^2/2 + 26.5*27 + 1 = 352.

Thus
the maximum possible value of an element in A is 352.