Question

If a ,b ,c, d and p are different real numbers such that:; (a2 +b2 +c2 +)p2 - 2(ab + bc +cd)p + (b2 +c2+ d2) <0 then show that a b c d are in GP

Answer 1

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Answer 2

Step-by-step explanation:

In the question,

a, b, c, d and P are real numbers.

Now, also,

$(a^{2}+b^{2}+c^{2})p^{2}-2(ab+bc+cd)p+(b^{2}+c^{2}+d^{2})<0.$

So,

On expanding and opening the brackets of the given equation we get,

$(a^{2}+b^{2}+c^{2})p^{2}-2(ab+bc+cd)p+(b^{2}+c^{2}+d^{2})\leq0\\a^{2}.p^{2}+b^{2}.p^{2}+c^{2}.p^{2}-2abp-2bcp-2cdp+b^{2}+c^{2}+d^{2}\leq0\\(a^{2}.p^{2}-2abp+b^{2})+(b^{2}.p^{2}-2bcp+c^{2})+(c^{2}.p^{2}-2cdp+d^{2})\leq0\\(ap-b)^{2}+(bp-c)^{2}+(cp-d)^{2}\leq 0$

So,

For the last equation to be possible the only case is when the equation is = 0.

So,

$(ap-b)^{2}+(bp-c)^{2}+(cp-d)^{2}= 0$

(ap - b) = 0

(bp - c) = 0

(cp - d) = 0

So,

$p=\frac{b}{a}\\and,\\p=\frac{c}{b}\\and,\\p=\frac{d}{c}$

On equating this we get,

$\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\\So,\\b^{2}=ac\\c^{2}=bd$

We know that, for the numbers a, b, c, d to be in G.P,

$\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\\So,\\b^{2}=ac\\  and,\\c^{2}=bd$

Therefore, we can from this say that the a, b, c, d are in G.P.