If a, b, c are in H.P.,then for all n in N (n | vedclass

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(B) Since $a, b, c$ are in $H.P.$,we have $b = \frac{2ac}{a+c}$.By the Power Mean Inequality,for $n > 1$ and $a 
eq c$,we have $\frac{a^n + c^n}{2} >

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$a^n + c^n > 2b^n$

Vedclass · Answer

(B) Since $a, b, c$ are in $H.P.$,we have $b = \frac{2ac}{a+c}$.By the Power Mean Inequality,for $n > 1$ and $a 
eq c$,we have $\frac{a^n + c^n}{2} > \left(\frac{a+c}{2}ight)^n$.Since $A.M. > H.M.$,we have $\frac{a+c}{2} > b$.Raising both sides to the power $n$ (where $n > 1$),we get $\left(\frac{a+c}{2}ight)^n > b^n$.Combining these inequalities,we get $\frac{a^n + c^n}{2} > \left(\frac{a+c}{2}ight)^n > b^n$.Therefore,$a^n + c^n > 2b^n$.

If $a, b, c$ are in $H.P.$,then for all $n \in N$ $(n > 1)$,the true statement is:

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