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  4. If a, b, c, d and p are distinct real numbers such that (a2 + b2 + c2) p2 - 2 (ab + bc + cd) p + (b2 + c2 + d2) le 0, then a, b, c, d

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Q. If $a, b, c, d$ and $p$ are distinct 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 $

IIT JEEIIT JEE 1987Sequences and Series

Solution:

Here, $ (a^2 + b^2 + c^2)\, p^2 - 2 (ab + bc + cd)\, p$
$ + (b^2 + c^2 + d^2) \le 0, $
$\, \Rightarrow (a^2 p^2 - 2abp + b^2) + (b^2 p^2 - 2bcp + c^2) $
$ + (c^2 p^2 - 2cdp + d^2) \le 0 $
$\, \Rightarrow (ap-b)^2 + (bp- c)^2 +(cp-d)^2 \le 0$
$ $ [since, sum of squares is never less than zero]
Since, each of the squares is zero.
$ \therefore (ap-b)^2 = (bp-c)^2 = (cp-d)^2 = 0 $
$ \Rightarrow p = \frac{b}{a} = \frac{c}{d} = \frac{d}{c}$
$ \therefore a, b, c, d$ are in GP.