Arithmetic Progression

Date: 12/19/96 at 19:14:26
From: Sushant
Subject: Arithmetic progression

I am stuck on two problems involving arithmetic progression:

1. If (b+c-a)/a, (c+a-b)/b and (a+b-c)/c are in arithmetic 
progression, show that 1/a, 1/b and 1/c are also in A.P.

2. If a+b+c is not equal to zero and (b+c)/a, (c+a)/b and (a+b)/c are 
in A.P., show that 1/a, 1/b and 1/c are also in A.P.

Date: 12/19/96 at 20:25:59
From: Doctor Pete
Subject: Re: Arithmetic progression

Hi,

(1) Take the difference of the first two terms, (b+c-a)/a - (c+a-b)/b, 
    and set it equal to the difference of the last two terms, 
    (c+a-b)/b - (a+b-c)/c.  Simplify.  This will give a relation 
    between 1/a, 1/b, and 1/c which indicates that their successive 
    differences will also be the same.  

    If a + b + c = 0, then the statement is false (prove it!).

(2) This follows directly from the first problem -- they are the same.

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 12/19/96 at 21:03:25
From: Sushant
Subject: Arithmetic progression

I tried that but I am not getting the answer. Please help.

Date: 12/21/96 at 03:15:40
From: Doctor Pete
Subject: Re: Arithmetic progression

1)  Let

     x1 = (b+c-a)/a = (b+c)/a - 1,
     x2 = (c+a-b)/b = (a+c)/b - 1,
     x3 = (a+b-c)/c = (a+b)/c - 1.

Then consider the successive differences

     x3 - x2 = (a+b)/c - 1 - (a+c)/b + 1 = (a+b)/c - (a+c)/b
             = (b(a+b) - c(a+c))/(bc) = (ab + b^2 - ac - c^2)/(bc)
             = (a(b-c) + b^2 - c^2)/(bc)
             = (a(b-c) + (b+c)(b-c)/(bc)
             = (a+b+c)(b-c)/(bc) ,
and
     x2 - x1 = (a+c)/b - 1 - (b+c)/a + 1
             = (a+b+c)(a-b)/(ab)

Since these must be equal if x1, x2, x3 are in arithmetic progression, 
it follows that:

     (a+b+c)(b-c)/(bc) = (a+b+c)(a-b)/(ab),

whereupon dividing by a+b+c (since it cannot be zero), we see:

     (b-c)/(bc) = (a-b)/(ab)

whereupon separating the terms in the numerator gives:

     1/c - 1/b = 1/b - 1/a

But if we consider the sequence y1 = 1/a, y2 = 1/b, and y3 = 1/c, this
implies that the successive differences y3 - y2 = y2 - y1, hence y1, 
y2, y3 are in also in arithmetic progression, and we are done.

Note that in order for the problem stated to be true, we must have a, 
b, c, and a+b+c nonzero.  It is clear that a, b, and c mustn't be 
zero, but to see why their sum must also be nonzero, observe that if 
it were zero, then:
 
x1 = (b+c-a)/a = (-2a)/a = -2 
x2 = (c+a-b)/b = (-2b)/b = -2
x3 = (a+b-c)/c = (-2c)/c = -2

So x1, x2, x3 form a trivial arithmetic sequence of common difference 
0 for *any* choice of a, b, c.  Hence the reciprocals 1/a, 1/b, 1/c 
need not be in arithmetic progression.

The answer to problem 2 is obviously a direct result of problem 1, for 
if a sequence x1, x2, x3 is arithmetic with common difference d, then 
(x1)+n, (x2)+n, (x3)+n is also arithmetic with common difference d, 
for any value of n.

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 12/21/96 at 22:38:41
From: Sushant
Subject: Re: Arithmetic progression

Thanks a lot. I am enlightened.