Progressive equations

$a,b,c$ are in Arithmetic Progression , so $a+c=2b.$ Since $a,b,c$ are non-zero, we can divide by $a,$ giving $1+\frac { c }{ a } =\frac { 2b }{ a }.$

Via Vieta's Formula , the product of the roots is $\frac { c }{ a } =\alpha \beta$ and the sum of the roots is $-\frac { b }{ a } =\alpha +\beta,$ so we have $1+\alpha \beta =-2(\alpha +\beta) \qquad (i).$

Since $\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } ,\alpha +\beta ,{ \alpha }^{ 2 }+{ \beta }^{ 2 }$ are in Geometric Progression , we have $(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } )({ \alpha }^{ 2 }+{ \beta }^{ 2 })={ (\alpha +\beta ) }^{ 2 }.$

This reduces to ${ \alpha }^{ 2 }+{ \beta }^{ 2 }={ (\alpha +\beta ) }\alpha \beta.$ From $(i),$ we can substitute $\alpha +\beta = -\frac{1+\alpha\beta}{2},$ giving

${ (\alpha }+\beta )^{ 2 }-2\alpha \beta =(\alpha +\beta )\alpha \beta \qquad \Rightarrow \qquad { { 4(1+\alpha \beta ) }^{ 2 } }-2\alpha \beta =-\frac { (1+\alpha \beta ) }{ 2 } \alpha \beta.$

Solving this quadratic for $\alpha \beta,$ we have $\alpha \beta =\frac { 1 }{ 3 } ,-1$ but $\alpha \beta =-1$ is not possible because we can't have $\alpha+\beta = 0$ with our geometric progression.

Thus, $\frac { a }{ c } =\frac { 1 }{ \alpha \beta } =3.$

\frac { \alpha +\beta }{ \alpha \beta } ,\quad \alpha +\beta \quad ,\quad and\quad { \alpha }^{ 2 }+{ \beta }^{ 2 }\quad in\quad GP\\ we\quad have\quad \\ by\quad property\quad of\quad roots\\ \alpha +\beta \quad =\quad -b/a\\ \alpha \beta \quad =\quad c/a\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }=\quad \frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \\ \\ now\quad as\quad they\quad are\quad in\quad GP\\ we\quad have\quad \\ (\frac { \alpha +\beta }{ \alpha \beta } )({ \alpha }^{ 2 }+{ \beta }^{ 2 })=\quad (\alpha +\beta )\^ 2\\ \\ now\quad putting\quad their\quad values\quad in\quad terms\quad of\quad a,b,c\quad \\ and\quad then\quad using\quad b\quad =\quad (a+c)/2\quad we\quad have\quad the\quad quadratic\quad interms\quad of\quad c/a\quad as\\ \\ { z }^{ 2 }-4z+3\quad where\quad z\quad =\quad c/a\\ solving\quad we\quad get\quad z=\quad 3\quad or\quad z=1,\quad as\quad a\quad and\quad c\quad distinct,\quad and\quad if\quad we\quad assume\quad a=\quad c\quad we\quad get\quad absurd\quad results\quad for\quad roots\\ we\quad have\quad a/c\quad =3