Geometric Progression...If a, b , c, d and p are different real numbers such that: (a^2 + b^2 + c^2)p^2 - 2(ab+bc+cd)p + (b^2 + c^2 + d^2)

Given that (a^2+b^2+c^)p^2-2(ab+bc+cd)p+(b^2+c^2+d^2)
< = 0. To prove that a,b,c and d are in Geometric
progression>

Solution:

We
observe that the left side of the inequality  could be written
like:

 (ap-b)^2 +(bp-c)^2+(cp-d)^2  the sum of which must
be > = 0 as each of these 3 terms is a  perfect
square..........(1)

But by the given condition
(ap-b)^2+(bp-c)^2+(cp-d)^2 <= 0........(2)

Therefore
the conditions (1) and(2) can be satisfying iff the sum, (ap-b)^2+(bp-c)62+(cp-d)^2 = 0 
which is possible  iff  each of the terms, (ap-b) =(bp-c) = (cp-d)  is equal to zero.
So,

ap-b = 0,  Or a/b =
p.

bp-c = 0. Or b/c = p.

cp-d
= o, Or a/d =p

Therefore, ab = b/c = c/d = p the common
ratio of the terms a, b,c and d.

So a,b,c,and d are in
geometric progression .