If $a, b, c, d$ and $p$ are different real numbers such that $\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+d^{2}\right)\, \leq \,0,$ then show that $a, b, c$ and $d$ are in $G.P.$
If $a, b, c, d$ and $p$ are different real numbers such that $\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+d^{2}\right)\, \leq \,0,$ then show that $a, b, c$ and $d$ are in $G.P.$
Given that
$\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+d^{2}\right) \,\leq \,0$ .........$(1)$
But $L.H.S.$
$=\left(a^{2} p^{2}-2 a b p+b^{2}\right)+\left(b^{2} p^{2}-2 b c p+c^{2}\right)+\left(c^{2} p^{2}-2 c d p+d^{2}\right)$
which gives $(a p-b)^{2}+(b p-c)^{2}+(c p-d)^{2}\, \geq \,0$ ..........$(2)$
Since the sum of squares of real numbers is non negative, therefore, from $(1)$ and $(2),$
we have, $(a p-b)^{2}+(b p-c)^{2}+(c p-d)^{2}=0$
or $a p-b=0, b p-c=0, c p-d=0$
This implies that $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=p$
Hence $a, b, c$ and $d$ are in $G.P.$
Similar Questions
Suppose four distinct positive numbers $a_1, a_2, a_3, a_4$ are in $G.P.$ Let $b_1=a_1, b_2=b_1+a_2, b_3=b_2+a_3$ and $b_4=b_3+a_4$.
$STATEMENT-1$ : The numbers $\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3, \mathrm{~b}_4$ are neither in $A.P$. nor in $G.P.$ and
$STATEMENT-2$ : The numbers $\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3, \mathrm{~b}_4$ are in $H.P.$
Suppose four distinct positive numbers $a_1, a_2, a_3, a_4$ are in $G.P.$ Let $b_1=a_1, b_2=b_1+a_2, b_3=b_2+a_3$ and $b_4=b_3+a_4$.
$STATEMENT-1$ : The numbers $\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3, \mathrm{~b}_4$ are neither in $A.P$. nor in $G.P.$ and
$STATEMENT-2$ : The numbers $\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3, \mathrm{~b}_4$ are in $H.P.$
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