If a, b, c, d and p are different real number | vedclass

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(B) Given the inequality: $(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \le 0$ $(i)$We can rewrite the expression on the left-hand side

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$G.P.$

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(B) Given the inequality: $(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \le 0$ $(i)$We can rewrite the expression on the left-hand side as:$(a^2p^2 - 2abp + b^2) + (b^2p^2 - 2bcp + c^2) + (c^2p^2 - 2cdp + d^2) \le 0$This simplifies to:$(ap - b)^2 + (bp - c)^2 + (cp - d)^2 \le 0$ $(ii)$Since the sum of squares of real numbers is always non-negative,the only way the sum can be less than or equal to $0$ is if each square term is exactly $0$:$(ap - b)^2 = 0, (bp - c)^2 = 0, (cp - d)^2 = 0$This implies:$ap = b, bp = c, cp = d$Therefore:$\frac{b}{a} = p, \frac{c}{b} = p, \frac{d}{c} = p$Since the ratio between consecutive terms is constant $(p)$,the sequence $a, b, c, d$ is in $G.P.$

If $a, b, c, d$ and $p$ are different real numbers such that $(a^2 + b^2 + c^2)p^2 - 2(ab + bc + cd)p + (b^2 + c^2 + d^2) \le 0$,then $a, b, c, d$ are in

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