Question

Find 'n' so that a^(n+2) + b^(n+2)/a^(n+1) + b^(n+1) is the A.M between a and b

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:\dfrac{ {a}^{n + 2} + {b}^{n + 2} }{ {a}^{n + 1} + {b}^{n + 1} } is \: the \: arithmetic\: mean \: between \:a \: and \: b$

So, it implies,

$\rm :\longmapsto\:\dfrac{ {a}^{n + 2} + {b}^{n + 2} }{ {a}^{n + 1} + {b}^{n + 1} } = \dfrac{a + b}{2}$

$\rm :\longmapsto\:2( {a}^{n + 2} + {b}^{n + 2}) = (a + b)( {a}^{n + 1} + {b}^{n + 1})$

$\rm :\longmapsto\:2{a}^{n + 2} + 2{b}^{n + 2}= {a}^{n + 2} + a{b}^{n + 1}+ b {a}^{n + 1} + {b}^{n + 2}$

$\rm :\longmapsto\:2{a}^{n + 2} + 2{b}^{n + 2} - {a}^{n + 2} - {b}^{n + 2} = a{b}^{n + 1}+ b {a}^{n + 1}$

$\rm :\longmapsto\:{a}^{n + 2} + {b}^{n + 2}= a{b}^{n + 1}+ b {a}^{n + 1}$

$\rm :\longmapsto\:{a}^{n + 2} - {ba}^{n + 1}= a{b}^{n + 1} - {b}^{n + 2}$

$\rm :\longmapsto\: {a}^{n + 1}(a - b) = {b}^{n + 1}(a - b)$

$\rm :\longmapsto\: {a}^{n + 1} = {b}^{n + 1}$

$\rm :\longmapsto\:\dfrac{ {a}^{n + 1} }{ {b}^{n + 1} } = 1$

$\rm :\longmapsto\: {\bigg(\dfrac{a}{b} \bigg) }^{n + 1} = {\bigg(\dfrac{a}{b} \bigg) }^{0}$

So, on comparing, we get

$\rm :\longmapsto\:n + 1 = 0$

$\bf\implies \:n = - \: 1$

Additional Information :-

1. Arithmetic mean > Geometric mean if observations are distinct.

2. Geometric mean between 2 positive numbers a and b is

$\sf \: \sqrt{ab}$

3. If n A.M's are inserted between two numbers a and b, then sum of n A.M's is n times the single AM between a and b.