Question

If a^n+b^n/a^n-1+b^n-1 is the A.M. between a and b then find the value of n

Answer 1

Step-by-step explanation:

We have,

$\tt{ \dfrac{ {a}^{n} + {b}^{n} }{ {a}^{n - 1} + {b}^{n - 1} } \,\,is\,\,the\, \,AM \,\,between\,\,a\,\, and\,\, b}$

So,

$\sf{ \dfrac{ {a}^{n} + {b}^{n} }{ {a}^{n - 1} + {b}^{n - 1} } = \dfrac{a + b}{2} }$

$\sf{ \implies 2{a}^{n} + 2{b}^{n} = ( {a}^{n - 1} + {b}^{n - 1} ) (a + b)}$

$\sf{ \implies 2{a}^{n} + 2{b}^{n} = {a}^{n} + ab ^{n - 1} + ba^{n - 1} + {b}^{n} }$

$\sf{ \implies {a}^{n} + {b}^{n} = ab ^{n - 1} + ba^{n - 1} }$

$\sf{ \implies {a}^{n} - ba^{n - 1} + {b}^{n} - ab ^{n - 1} = 0 }$

$\sf{ \implies {a}^{n - 1}(a - b) - {b}^{n - 1}(a - b ) = 0 }$

$\sf{ \implies( {a}^{n - 1}- {b}^{n - 1})(a - b ) = 0 }$

$\rm{ \color{orange}Since \: \: a \: \: and \: \: b \: \: are \: \: distinct \: , so}$

$\sf{ \implies{a}^{n - 1} - {b}^{n - 1} = 0 }$

$\sf{ \implies{a}^{n - 1} = {b}^{n - 1} }$

$\sf{ \implies{ \bigg( \dfrac{a}{b} \bigg) }^{n - 1} = 1 }$

$\sf{ \implies \: n = 1 }$