Question

C^n+1 + d^n+1 / C^n +d^n i sthe arithmetic mean of C and d, then n is _____?

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Answer 1

Answer:

We know that, AM of any two numbers c and d given by,

$\frac{c + d}{2}$

Now,

$\frac{c + d}{2} = \frac{ {c}^{n + 1} + {d}^{n + 1} }{ {c}^{n} + {d}^{n} }$

$\implies(c + d) ({c}^{n} + {d}^{n} ) = 2({c}^{n + 1} + {d}^{n + 1} )$

$\implies({c}^{n + 1} + c {d}^{n} + d{c}^{n} + {d}^{n + 1} ) = 2({c}^{n + 1} + {d}^{n + 1} )$

$\implies( c {d}^{n} + d{c}^{n} ) = ({c}^{n + 1} + {d}^{n + 1} )$

$\implies {c}^{n + 1} - d{c}^{n}+ {d}^{n + 1} - c {d}^{n}= 0$

$\implies {c}^{n } (c- d)+ {d}^{n}(d - c)= 0$

$\implies {c}^{n } (c- d) - {d}^{n}( c - d)= 0$

$\implies ({c}^{n } - {d}^{n})( c - d)= 0$

$\implies {c}^{n } - {d}^{n}= 0$

$\implies { (\frac{c}{d})}^{n } = 1$

$\implies \: n = 0$