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

Question

Dr. Abubacker · Accepted Answer

(b)
We have,
$\left( a^{2} + b^{2} + c^{2} ight)p^{2} - 2(ab + bc + cd)p + (b^{2} + c^{2} + d^{2}) \leq 0$ …(i)
LHS$\  = \left( a^{2}p^{2} - 2abp + b^{2} ight) + \left( b^{2}p^{2} - 2bcp + c^{2} ight) + (c^{2}p^{2} - 2cdp + d^{2})$ …(ii)
$= \left( ap - b^{2} ight) + \left( bp - c^{2} ight) + (cp - d^{2}) \geq 0$
Since, the sum of square of real number is non-negative
From Eqs.(i) and (ii), we get
$(ap - b)^{2} + (bp - c)^{2} + (cp - d)^{2} = 0$
$\Rightarrow \ \ ap - b = 0 = bp - c = cp - d$
$\Rightarrow \ \frac{b}{a} = \frac{c}{d} = \frac{d}{c} = p$
$	herefore\ a,\ b,\ c,\ d$ are in GP.

Q. If $a,\ b,\ c,\ d\ and\ p$ are different real numbers such that $\left( a^{2} + b^{2} + c^{2} \right)\ p^{2} - 2(ab + bc + cd)\ \ p + \left( b^{2} + c^{2} + d^{2} \right) \leq 0,then\ a,\ b,\ c,\ d$ are in

Explanation

Reviewed by Dr. Abubacker

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