tmua.me Medium Set D Paper 2 · Question 19

Let $n\geq3$ , and let $a_1<a_2<\cdots<a_n$ be real numbers. Among all the positive differences $a_j-a_i$ , where $i<j$ , there are exactly $n-1$ distinct values.

The following incomplete proof shows that $a_1,\ldots,a_n$ form an arithmetic progression.

Consider the list

a_n-a_1>a_n-a_2>\cdots>a_n-a_{n-1}.

[Missing argument]

This establishes

a_{n-1}-a_i=a_n-a_{i+1} \qquad(1\leq i\leq n-2).

Rearranging shows that each of the first $n-2$ consecutive gaps equals the last gap, $a_n-a_{n-1}$ , proving the result.

Which argument correctly fills the gap without assuming the result to be proved?

AThe first list contains

n-1

entries, and there are

n-1

consecutive gaps. Thus the consecutive gaps exhaust this list. Comparing them in order gives the required equalities.

BFor

1\leq i\leq n-2

, we have

a_{n-1}-a_i<a_n-a_i

. Strict ordering alone then forces the left-hand value to be the next smaller entry of the first list.

CSubtract

a_n-a_{n-1}

from every entry of the first list. Each result is a positive difference already in the list, so the list is unchanged. Matching entries gives the required equalities.

DAll positive differences occur in the first list. The

n-2

distinct differences

a_{n-1}-a_i

with

1\leq i\leq n-2

are smaller than its largest entry, so they fill the other

n-2

entries. Match these two decreasing lists.

EWrite

a_i=a_1+(i-1)d

, where

d>0

. The first list is

(n-1)d,(n-2)d,\ldots,d

, and the differences from

a_{n-1}

are

(n-2)d,\ldots,d

. Comparing these lists gives the required equalities.

examday

tmua.me Medium Set D Paper 2 · Q19