Answer
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Hint: In order to solve the problem use the property of the discriminant of the quadratic equation when the roots of the quadratic equation are the same. Equate discriminant to be 0 in order to find the pattern between a, b and c.
Complete step-by-step answer:
Given quadratic equation: $a\left( {b - c} \right){x^2} + b\left( {c - a} \right)x + c\left( {a - b} \right) = 0$
As we know that for the general quadratic equation of the form $m{x^2} + nx + s = 0$ having some roots p and q. Whenever the roots of the quadratic equation are the same the discriminant is 0. Given by:
$D = {n^2} - 4ms = 0$
So comparing the coefficients of a given quadratic equation with the general form and making the discriminant 0. We get:
$
\because D = {n^2} - 4ms = 0 \\
\Rightarrow D = {\left( {b\left( {c - a} \right)} \right)^2} - 4\left( {a\left( {b - c} \right)} \right)\left( {c\left( {a - b} \right)} \right) = 0 \\
$
Now let us simplify the given equation.
$
\Rightarrow D = {b^2}{\left( {c - a} \right)^2} - 4ac\left( {b - c} \right)\left( {a - b} \right) = 0 \\
\Rightarrow D = {b^2}\left( {{c^2} + {a^2} - 2ac} \right) - 4ac\left( {ab - ac - {b^2} + bc} \right) = 0{\text{ }}\left[ {\because {{\left( {x - y} \right)}^2} = {x^2} + {y^2} - 2xy} \right] \\
$
Let us separate the similar terms:
$
\Rightarrow D = {b^2}\left( {{c^2} + {a^2} - 2ac + 4ac} \right) + 4{a^2}{c^2} - 4abc\left( {c + a} \right) = 0 \\
\Rightarrow D = {b^2}\left( {{c^2} + {a^2} + 2ac} \right) + {\left( {2ac} \right)^2} - 2\left( {2ac} \right)b\left( {c + a} \right) = 0 \\
\Rightarrow D = {b^2}{\left( {c + a} \right)^2} + {\left( {2ac} \right)^2} - 2\left( {2ac} \right)b\left( {c + a} \right) = 0{\text{ }}\left[ {\because {{\left( {x - y} \right)}^2} = {x^2} + {y^2} - 2xy} \right] \\
\Rightarrow D = {\left( {b\left( {c + a} \right)} \right)^2} + {\left( {2ac} \right)^2} - 2\left( {2ac} \right)b\left( {c + a} \right) = 0 \\
$
Now let us make the square of the terms using the algebraic identity
$ \Rightarrow D = {\left( {b\left( {c + a} \right) - 2ac} \right)^2} = 0{\text{ }}\left[ {\because {{\left( {x - y} \right)}^2} = {x^2} + {y^2} - 2xy} \right]$
Now let us find the relation between the variable a, b and c
\[
\Rightarrow b\left( {c + a} \right) - 2ac = 0 \\
\Rightarrow b\left( {c + a} \right) = 2ac \\
\Rightarrow b = \dfrac{{2ac}}{{\left( {c + a} \right)}} \\
\]
As we know that the relation received in the above equation is the formula for the Harmonic mean of terms.
Hence, terms a, b and c are in Harmonic progression as they satisfy the relation for harmonic mean.
So, option C is the correct option.
Note: Students must remember the formula for discriminant of quadratic equation. Also students must remember the characteristics of discriminants for different types of roots. Discriminant is zero for equal roots. Discriminant is positive for real and unequal roots. And discriminant is negative for imaginary roots.
Complete step-by-step answer:
Given quadratic equation: $a\left( {b - c} \right){x^2} + b\left( {c - a} \right)x + c\left( {a - b} \right) = 0$
As we know that for the general quadratic equation of the form $m{x^2} + nx + s = 0$ having some roots p and q. Whenever the roots of the quadratic equation are the same the discriminant is 0. Given by:
$D = {n^2} - 4ms = 0$
So comparing the coefficients of a given quadratic equation with the general form and making the discriminant 0. We get:
$
\because D = {n^2} - 4ms = 0 \\
\Rightarrow D = {\left( {b\left( {c - a} \right)} \right)^2} - 4\left( {a\left( {b - c} \right)} \right)\left( {c\left( {a - b} \right)} \right) = 0 \\
$
Now let us simplify the given equation.
$
\Rightarrow D = {b^2}{\left( {c - a} \right)^2} - 4ac\left( {b - c} \right)\left( {a - b} \right) = 0 \\
\Rightarrow D = {b^2}\left( {{c^2} + {a^2} - 2ac} \right) - 4ac\left( {ab - ac - {b^2} + bc} \right) = 0{\text{ }}\left[ {\because {{\left( {x - y} \right)}^2} = {x^2} + {y^2} - 2xy} \right] \\
$
Let us separate the similar terms:
$
\Rightarrow D = {b^2}\left( {{c^2} + {a^2} - 2ac + 4ac} \right) + 4{a^2}{c^2} - 4abc\left( {c + a} \right) = 0 \\
\Rightarrow D = {b^2}\left( {{c^2} + {a^2} + 2ac} \right) + {\left( {2ac} \right)^2} - 2\left( {2ac} \right)b\left( {c + a} \right) = 0 \\
\Rightarrow D = {b^2}{\left( {c + a} \right)^2} + {\left( {2ac} \right)^2} - 2\left( {2ac} \right)b\left( {c + a} \right) = 0{\text{ }}\left[ {\because {{\left( {x - y} \right)}^2} = {x^2} + {y^2} - 2xy} \right] \\
\Rightarrow D = {\left( {b\left( {c + a} \right)} \right)^2} + {\left( {2ac} \right)^2} - 2\left( {2ac} \right)b\left( {c + a} \right) = 0 \\
$
Now let us make the square of the terms using the algebraic identity
$ \Rightarrow D = {\left( {b\left( {c + a} \right) - 2ac} \right)^2} = 0{\text{ }}\left[ {\because {{\left( {x - y} \right)}^2} = {x^2} + {y^2} - 2xy} \right]$
Now let us find the relation between the variable a, b and c
\[
\Rightarrow b\left( {c + a} \right) - 2ac = 0 \\
\Rightarrow b\left( {c + a} \right) = 2ac \\
\Rightarrow b = \dfrac{{2ac}}{{\left( {c + a} \right)}} \\
\]
As we know that the relation received in the above equation is the formula for the Harmonic mean of terms.
Hence, terms a, b and c are in Harmonic progression as they satisfy the relation for harmonic mean.
So, option C is the correct option.
Note: Students must remember the formula for discriminant of quadratic equation. Also students must remember the characteristics of discriminants for different types of roots. Discriminant is zero for equal roots. Discriminant is positive for real and unequal roots. And discriminant is negative for imaginary roots.
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