# 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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Answer

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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.

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.

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