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

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Question

(b) As given, $({a^2} + {b^2} + {c^2}){p^2} - 2(ab + bc + cd)p$

$ + ({b^2} + {c^2} + {d^2}) \le 0$ &hellip;.. $(i)$

But $L.H.S.$

$ = ({a^2}{p

Vedclass · Accepted Answer

$G.P.$

Vedclass · Answer

(b) As given, $({a^2} + {b^2} + {c^2}){p^2} - 2(ab + bc + cd)p$

$ + ({b^2} + {c^2} + {d^2}) \le 0$ &hellip;.. $(i)$

But $L.H.S.$

$ = ({a^2}{p^2} - 2abp + {b^2}) + ({b^2}{p^2} - 2bcp + {c^2})$

$ + ({c^2}{p^2} - 2cdp + {d^2})$ $ = {(ap - b)^2} + {(bp - c)^2} + {(cp - d)^2} \ge 0$ &hellip;..$(ii)$

Since the sum of squares of real numbers is non-negative.

Therefore from $(i)$ and $(ii)$

$ \Rightarrow $${(ap - b)^2} + {(bp - c)^2} + {(cp - d)^2} = 0$

$ \Rightarrow $$ap - b = 0 = bp - c = cp - d \Rightarrow \frac{b}{a} = \frac{c}{b} = \frac{d}{c} = p$

$	herefore $$a,\;b,\;c,\;d$ are 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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