p,p2,p3 and sequence?

Question

A sequence P1,P2,P3......Pn of real numbers is defined such that P1=0; |P2|=|P1+1|;|P3|=|P2+1|;....|Pn|=|Pn-1+1|.

If n is an even natural number, the minimum possible value of (P1+P2+P3+...Pn)/n is

(1)-3/2
(2)-1
(3)-1/2
(4)0

Answer 1

The answer is (3) -1/2. The logic behind this is as follows:

In order to minimize the fraction, you want to minimize the numerator, since you can't control the denominator. The given recursive definition shows that one way to do this to, make each term as small as possible. Since P1=0 and |P2|=|P1+1|, the minimum value that |P2| can be is -1. From here, |P3|=|P2+1| = |-1+1|=|0|=0 and thus P3 is 0. Using the same logic, we see that P4 = -1, P5=0. Thus the sequence that minimizes each term is an alternating series of 0's and -1's. If n is even, exactly 1/2 of the terms will equal -1 and the other half will equal 0, so (P1+P2+P3+...+Pn)/n = (-n/2)/n = -1/2.

Answer 2

Here p1=0, |p2|=|p1+1|...
Hence p2=+/-1
p3=+/-2...

Since they are in sequence

The answer will be the smallest if the values of p are Negative

Let all P be Negative.

Then (P1 + P2 ...Pn)/n = {0-1-2-...-(n-1)}/n ...1

this can be written in a different way

={(n-1)-...3-2-1-0}/n ...2

Adding 1 and 2

2(P1+P2 + ....Pn)/n = {0-1-2...-(n-1)}/n+
{(n-1)-...2-1-0}/n

={0-(n-1) -1-(n-2)... -(n-2)-1-(n-1)-0)}/n

={(-n+1)(-n+1)...n times}/n

=(1-n)n/n

=(1-n)

Dividing both sides by 2

(P1+P2 + ....Pn)/n =(1-n)2

Now n is a even positive natural number

The smallest value it could hold is when n = 2

(1-n)2 = (1-2)/2

= -1/2


Answer is -1/2