Let a, b, c, d and p be any non-zero distinct | vedclass

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(C) Given the equation: $(a^{2}+b^{2}+c^{2}) p^{2} - 2(ab+bc+cd) p + (b^{2}+c^{2}+d^{2}) = 0$. Rearranging the terms,we get: $(a^{2}p^{2} - 2abp + b^{

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$a, b, c, d$ are in $G.P.$

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(C) Given the equation: $(a^{2}+b^{2}+c^{2}) p^{2} - 2(ab+bc+cd) p + (b^{2}+c^{2}+d^{2}) = 0$. Rearranging the terms,we get: $(a^{2}p^{2} - 2abp + b^{2}) + (b^{2}p^{2} - 2bcp + c^{2}) + (c^{2}p^{2} - 2cdp + d^{2}) = 0$. This can be written as: $(ap - b)^{2} + (bp - c)^{2} + (cp - d)^{2} = 0$. Since $a, b, c, d, p$ are real numbers,the sum of squares is zero only if each term is zero: $ap - b = 0 \Rightarrow p = \frac{b}{a}$ $bp - c = 0 \Rightarrow p = \frac{c}{b}$ $cp - d = 0 \Rightarrow p = \frac{d}{c}$ Thus,$\frac{b}{a} = \frac{c}{b} = \frac{d}{c} = p$. This implies that $a, b, c, d$ are in $G.P.$

Let $a, b, c, d$ and $p$ be any non-zero distinct 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:

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