Question

If $a,b,c$are non zero real numbers such that $3(a^2+b^2+c^2+1)=2(a+b+c+ab+bc+ca),$ then$a,b,c$ are in

Answer 1

Finding the relation of terms $a,b,c$:

Given $a,b,c$ are non zero real numbers and

$3(a^2+b^2+c^2+1)=2(a+b+c+ab+bc+ca)$

$\Rightarrow$ $3a^2+3b^2+3c^2+3-2a-2b-2c-2ab-2bc-2ac=0$

$\Rightarrow$$(a^2-2a)+(b^2-2b)+(c^2-2c)+(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)+1+1+1=0$

$\Rightarrow$$(a^2-2a+1)+(b^2-2b+1)+(c^2-2c+1)+(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)=0$

$\Rightarrow$ ${(a-1)}^2+{(b-1)}^2+{(c-1)}^2+{(a-b)}^2+{(b-c)}^2+{(c-a)}^2=0$

$\Rightarrow$ $(a-1)=0,(b-1)=0,(c-1)=0$

$\Rightarrow$ $a=1,b=1,c=1$

Here $2b=a+c$

So condition for $A.P$ is satisfied and

$b^2=ac$, So condition for $G.P$ also satisfied and

$\frac{2}{b}=\frac{(a+c)}{ac}$ So condition for $H.P$ also satisfied

Therefore the terms $a,b,c$ are in $A.P,G.P,H.P$

Hence the terms $\mathit{a}\mathbf{,}\mathit{b}\mathbf{,}\mathit{c}\mathbf{}$ are in $\mathit{A}\mathbf{.}\mathit{P}\mathbf{}\mathbf{,}\mathit{G}\mathbf{.}\mathit{P}\mathbf{}\mathbf{,}\mathit{H}\mathbf{.}\mathit{P}\mathbf{.}$