Correct Answer - Option 4 : harmonic progression
Concept:
Let us consider the standard form of a quadratic equation, ax2 + bx + c =0
Where a, b and c are constants with a ≠ 0
Let α and β be the two roots of the above quadratic equation.
The sum of the roots of a quadratic equation are:
\({\rm{\alpha }} + {\rm{\beta }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient\;of\;x}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}\)
The product of the roots is given by:
\({\rm{\alpha \beta }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant\;term}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}\)
Calculation:
We have, ax2 + bx + c =0
Let α and β be the two roots of the above quadratic equation.
The sum of the roots of a quadratic equation are: \({\rm{\alpha }} + {\rm{\beta }} = - \frac{{\rm{b}}}{{\rm{a}}}\)
The product of the roots is given by: \({\rm{\alpha \beta }} = \frac{{\rm{c}}}{{\rm{a}}}\)
Given that the sum of the roots of the quadratic equation ax2 + bx + c = 0 is equal to the sum of the squares of their reciprocals.
\(\alpha + \beta = \frac{1}{{{\alpha ^2}}} + \frac{1}{{{\beta ^2}}}\)
\( \Rightarrow \alpha + \beta = \frac{{{\alpha ^2} + {\beta ^2}}}{{{\alpha ^2}{\beta ^2}}} = \frac{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta }}{{{{\left( {\alpha \beta } \right)}^2}}}\)
\( \Rightarrow - \frac{b}{a} = \frac{{\frac{{{b^2}}}{{{a^2}}} - \frac{{2c}}{a}}}{{\frac{{{c^2}}}{{{a^2}}}}}\)
\( \Rightarrow - \frac{b}{a} = \frac{{{b^2} - 2ac}}{{{c^2}}}\)
\( \Rightarrow \frac{b}{c} + \frac{c}{a} = \frac{{2a}}{b}\)
From the above equation, it is clear that \(\frac{a}{c},\frac{b}{a}\) and \(\frac{c}{b}\) are in harmonic progression.
