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Question: 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...

If a,b,ca,b,c are in A.P., a2,b2,c2{a^2},{b^2},{c^2} are in H.P., then prove that either a=b=ca = b = c or a,b,−c2a,b, - \dfrac{c}{2} form a G.P.

Explanation

Solution

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,zx,y,z are 2y=x+z , y2=xz , and 2y=1x+1z2y = 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,ca,b,c are in A.P.
i.e. 2b=a+c2b = a + c
Which can be converted into b−a=c−b............................(1)b - a = c - b............................\left( 1 \right)
And a2,b2,c2{a^2},{b^2},{c^2} are in H.P.
i.e. 2b2=1a2+1c2\dfrac{2}{{{b^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{c^2}}}
Which can be written as

1b2−1a2=1c2−1b2  a2−b2a2b2=b2−c2b2c2 \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 p2−q2=(p+q)(p−q){p^2} - {q^2} = \left( {p + q} \right)\left( {p - q} \right)we can write as

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.