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
if $x = \,\frac{4}{3}\, - \,\frac{{4x}}{9}\, + \,\,\frac{{4{x^2}}}{{27}}\, - \,\,.....\,\infty $ , then $x$ is equal to
if $x = \,\frac{4}{3}\, - \,\frac{{4x}}{9}\, + \,\,\frac{{4{x^2}}}{{27}}\, - \,\,.....\,\infty $ , then $x$ is equal to
Let $x _{1}, x _{2}, x _{3}, \ldots ., x _{20}$ be in geometric progression with $x_{1}=3$ and the common ration $\frac{1}{2}$. A new data is constructed replacing each $x_{i}$ by $\left(x_{i}-i\right)^{2}$. If $\bar{x}$ is the mean of new data, then the greatest integer less than or equal to $\bar{x}$ is $.....$
Let $x _{1}, x _{2}, x _{3}, \ldots ., x _{20}$ be in geometric progression with $x_{1}=3$ and the common ration $\frac{1}{2}$. A new data is constructed replacing each $x_{i}$ by $\left(x_{i}-i\right)^{2}$. If $\bar{x}$ is the mean of new data, then the greatest integer less than or equal to $\bar{x}$ is $.....$
- [JEE MAIN 2022]
The third term of a $G.P.$ is the square of first term. If the second term is $8$, then the ${6^{th}}$ term is
The third term of a $G.P.$ is the square of first term. If the second term is $8$, then the ${6^{th}}$ term is
If $x > 1,\;y > 1,z > 1$ are in $G.P.$, then $\frac{1}{{1 + {\rm{In}}\,x}},\;\frac{1}{{1 + {\rm{In}}\,y}},$ $\;\frac{1}{{1 + {\rm{In}}\,z}}$ are in
If $x > 1,\;y > 1,z > 1$ are in $G.P.$, then $\frac{1}{{1 + {\rm{In}}\,x}},\;\frac{1}{{1 + {\rm{In}}\,y}},$ $\;\frac{1}{{1 + {\rm{In}}\,z}}$ are in
- [IIT 1998]
The remainder when the polynomial $1+x^2+x^4+x^6+\ldots+x^{22}$ is divided by $1+x+x^2+x^3+\ldots+x^{11}$ is
The remainder when the polynomial $1+x^2+x^4+x^6+\ldots+x^{22}$ is divided by $1+x+x^2+x^3+\ldots+x^{11}$ is
- [KVPY 2016]