If $$a,b,c$$ are in H.P. and they are distinct and positive, then prove that $${a^n} + {c^n} 2{b^n}$$

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If $$a,b,c$$ are in H.P. and they are distinct and positive, then prove that $${a^n} + {c^n} > 2{b^n}$$

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Hint: First of all, find an equation between the numbers $$a,b,c$$ which is in H.P. Then find the geometric mean and harmonic mean of $$a$$ and $$c$$. Use A.M > G.M > H.P to prove the given statement. So, use this concept to reach the solution of the given problem. Complete step-by-step answer: Given that $$a,b,c$$ are in H.P. and they are distinct and positive.So, we have $$\dfrac{1}{a} + \dfrac{1}{c} = \dfrac{2}{b} \Rightarrow b = \dfrac{{2ac}}{{a + c}}$$Consider geometric mean of $$a$$ and $$c$$ = $$\sqrt {ac} $$Harmonic mean of $$a$$ and $$c$$ = $$\dfrac{{2ac}}{{a + c}}$$We know that G.M > H.PSo, we have $$   \Rightarrow \sqrt {ac}  > \dfrac{{2ac}}{{a + c}}   \   \Rightarrow \sqrt {ac}  > b\,{\text{                                       }}\left[ {\because \dfrac{{2ac}}{{a + c}} = b} \right]   \ $$Raising its power to $$n$$ on both sides, we get$$   \Rightarrow {\left( {\sqrt {ac} } \right)^n} > {b^n}   \   \Rightarrow {\left( {ac} \right)^{\dfrac{n}{2}}} > {b^n}....................................\left( 1 \right)   \ $$Let $${a^n}$$ and $${c^n}$$ be two numbers. Since $$a,c$$ are positive and distinct, $${a^n}$$ and $${c^n}$$ are also positive. We know that for two numbers A.M > G.M > H.MNow, Arithmetic mean (A.M) of $${a^n}$$ and $${c^n}$$ = $$\dfrac{{{a^n} + {c^n}}}{2}$$Geometric mean (G.M) of $${a^n}$$ and $${c^n}$$ = $$\sqrt {{a^n}{c^n}}  = {\left( {ac} \right)^{\dfrac{n}{2}}}$$Harmonic mean of $${a^n}$$ and $${c^n}$$ = $$\dfrac{{2{a^n}{c^n}}}{{{a^n} + {c^n}}} = \dfrac{{2{{\left( {ac} \right)}^{\dfrac{n}{2}}}}}{{{a^n} + {c^n}}}$$Since A.M > G.M$$   \Rightarrow \dfrac{{{a^n} + {c^n}}}{2} > {\left( {ac} \right)^{\dfrac{n}{2}}}   \   \Rightarrow {a^n} + {c^n} > 2{\left( {ac} \right)^{\dfrac{n}{2}}}   \  \therefore {a^n} + {c^n} > 2{b^n}{\text{                                                    }}\left[ {\because {\text{equation }}\left( 1 \right)} \right]   \ $$Hence proved that $${a^n} + {c^n} > 2{b^n}$$Note: Three numbers $$x,y,z$$ are said to be in Harmonic progression if it satisfies the condition $$\dfrac{1}{x} + \dfrac{1}{z} = \dfrac{2}{y}$$.  Arithmetic mean, geometric mean and harmonic mean of two numbers $$x$$ and $$y$$ is given by $$\dfrac{{x + y}}{2},\sqrt {xy} ,\dfrac{{2xy}}{{x + y}}$$ respectively.

If \[a,b,c\] are in H.P. and they are distinct and positive, then prove that \[{a^n} + {c^n} > 2{b^n}\]

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If \[a,b,c\] are in H.P. and they are distinct and positive, then prove that \[{a^n} + {c^n} > 2{b^n}\]

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