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Local Maxima

If a, b, c, d...

Question

If $a, b, c, d & p$ are distinct real numbers such that $(a^{2} + b^{2} + c^{2}) p^{2} - 2 (a b + b c + c d) p + (b^{2} + c^{2} + d^{2}) \leq 0,$ Then $a, b, c & d$

$a r e i n a r i t h m e t i c p r o g r e s s i o n$

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$a r e i n g e o m e t r i c p r o g r e s s i o n$

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$a r e i n h a r m o n i c p r o g r e s s i o n$

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$s a t i s f y a b = c d$

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Solution

The correct option is B
$a r e i n g e o m e t r i c p r o g r e s s i o n$

Step1. Expressing the given data:
Given, $(a^{2} + b^{2} + c^{2}) p^{2} - 2 (a b + b c + c d) p + (b^{2} + c^{2} + d^{2}) \leq 0,$
$a, b, c, d & p$ are distinct real numbers
Step2. Finding the progression.
Now,
$\begin{array}{rcl} (a^{2} p^{2} + b^{2} p^{2} + p^{2} c^{2}) - 2 (a b p + b c p + c d p) + (b^{2} + c^{2} + d^{2}) & \leq & 0 \\ (a^{2} p^{2} + b^{2} + - 2 a b p) + (b^{2} p^{2} - 2 b c p + c^{2}) + (c^{2} p^{2} - 2 c d p + d^{2}) & \leq & 0 \\ {(a p - b)}^{2} + {(b p - c)}^{2} + {(c p - d)}^{2} & \leq & 0 \end{array}$
Sum of the square cannot be negative.
$\begin{array}{rcl} s o, (a p - b) & = & 0 \\ a p & = & b \dots (i) \\ (b p - c) & = & 0 \\ b p & = & c \dots (i i) \\ c p - d & = & 0 \\ c p & = & d \dots (i i i) \end{array}$
From $(i) & (i i)$
$\begin{array}{rcl} \frac{b}{a} & = & \frac{c}{b} \\ \Rightarrow & b^{2} = a c \end{array}$
From $(i i) & (i i i)$
$\begin{array}{rcl} \frac{c}{b} & = & \frac{d}{c} \\ \Rightarrow & c^{2} = b d \end{array}$

Therefore, $a, b, c & d$ are in Geometric Progression.

Hence correct option is $(B)$

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