# 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) <= 0, then show that a, b , c , d are in G.P.

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### 2 Answers

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 .

We have:

(a^2 + b^2 + c^2)p^2 - 2(ab+bc+cd)p + (b^2 + c^2 + d^2) <=0

This can be rewritten as:

a^2*p^2 + b^2*p^2 + c^2*p^2 - 2abp-2bcp-2cdp + b^2 + c^2 + d^2 <=0

Arranging the terms:

[a^2*p^2- 2abp+ b^2] + [b^2*p^2-2bcp+ c^2 ]+ [c^2*p^2 -2cdp + d^2] <=0

(ap-b)^2+(bp-c)^2+(cp-d)^2<=0

As the three terms are being squared they cannot be less than 0, therefore they are equal to 0.

So ap-b=0, bp-c=0 and cp-d=0

=>b/a=p, c/b=p, d/c=p

Therefore, we have a, b, c and d are in G.P. as each term has the forms the same ratio with the preceding term.