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# If $a,b,c$ are in A.P., ${a^2},{b^2},{c^2}$ are in H.P., then prove that either $a = b = c$ or $a,b, -\dfrac{c}{2}$ form a G.P.

Last updated date: 22nd Mar 2023
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Hint: Use properties of AP and HP on given conditions to get the desired result, that is form two equations using AM and HM and simplify.

Step by step solution:
We know that the conditions to be in A.P., G.P., and H.P. of the terms $x,y,z$ are $2y = x + z{\text{ }},{\text{ }}{y^2} = xz{\text{ }},{\text{ and }}\dfrac{2}{y} = \dfrac{1}{x} + \dfrac{1}{z}$ respectively.
So, by using the above formulae we can solve this problem easily.
Given $a,b,c$ are in A.P.
i.e. $2b = a + c$
Which can be converted into $b - a = c - b............................\left( 1 \right)$
And ${a^2},{b^2},{c^2}$ are in H.P.
i.e. $\dfrac{2}{{{b^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{c^2}}}$
Which can be written as
$\dfrac{1}{{{b^2}}} - \dfrac{1}{{{a^2}}} = \dfrac{1}{{{c^2}}} - \dfrac{1}{{{b^2}}} \\ \\ \dfrac{{{a^2} - {b^2}}}{{{a^2}{b^2}}} = \dfrac{{{b^2} - {c^2}}}{{{b^2}{c^2}}} \\$
By using the formula ${p^2} - {q^2} = \left( {p + q} \right)\left( {p - q} \right)$we can write as
$\dfrac{{\left( {a - b} \right)\left( {a + b} \right)}}{{{a^2}{b^2}}} = \dfrac{{\left( {b - c} \right)\left( {b + c} \right)}}{{{b^2}{c^2}}}.........................\left( 2 \right)$
From equations (1) and (2) we get
$\dfrac{{\left( {a + b} \right)}}{{{a^2}}} = \dfrac{{\left( {b + c} \right)}}{{{c^2}}} \\ \\ a{c^2} + b{c^2} = b{a^2} + c{a^2} \\ \\ a{c^2} - c{a^2} + b{c^2} - b{a^2} = 0 \\ \\ ac\left( {c - a} \right) + b\left( {{c^2} - {a^2}} \right) = 0 \\ \\ ac\left( {c - a} \right) + b\left( {c - a} \right)\left( {c + a} \right) = 0 \\ \\ \left[ {ac + b\left( {c + a} \right)} \right]\left( {c - a} \right) = 0 \\$
Either $ac + b\left( {c + a} \right) = 0$ or $\left( {c - a} \right) = 0$
If $\left( {c - a} \right) = 0$ then $a = c...............................(3)$
Now consider $ac + b\left( {c + a} \right) = 0$
$\Rightarrow ac + b\left( {2b} \right) = 0{\text{ [}}\because 2b = a + c] \\ \Rightarrow ac + 2{b^2} = 0 \\ \Rightarrow 2{b^2} = - ac \\ \Rightarrow {b^2} = - \dfrac{{ac}}{2} \\$
Therefore $a,b,\dfrac{{ - c}}{2}$ are in G.P.
We have $2b = a + c$ and $a = c$
So,
$2b = a + a \\ 2b = 2a \\ \therefore a = b \\$
Therefore $a = b = c$
Hence proved that $a = b = c$ or $a,b,\dfrac{{ - c}}{2}$ are in G.P.

Note: Harmonic Progression is the reciprocal of the values of the terms in Arithmetic Progression.
And equate each equation to zero to know all the values.