Question

If( a^n+b^n)/a^n-1 +b^n-1 is the arithematic mean between a and b. Then find n.

Answer 1

Solution:-

Arithmetic mean of a and b is=[a+b]/2

So that,

[a+b]/2=[a^n+1+b ^n+1]/a^n+b^n

[a+b]/2=a^n*a+b^n*b/a^n+b^n

(ax+y=ax*ay)

Then by cross multiplication we get

a^na+a^nb+b^na+b^nb=2a^na+2b^nb

a^nb+b^na=a^na+b^nb

a^nb-a^na=b^nb-b^na

a^n(b-a)=b^n(b-a)

a^n=b^n

a and b can only be = if n=0 (x^n=1)

[Therefore, n=0]

Answer 2

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$




$\mathsf{Arithmetic \: between \: a \: and \: b \: = \: \dfrac{a \: + \: b}{2}}$




$\underline{\textsf{According to question : }} \\ \\ \sf \implies \dfrac{a^n \: + \: b^n}{a^{(n \: - \: 1 )} \: + \: b^{( n \: - \: 1 )} } \: = \: \dfrac{a \: + \: b}{2}$



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




$\sf \implies {a}^{(n \: - 1)} \cancel{ \{a \: - \: b \}}\: = \: b^{(n \: - \: 1)} \cancel{ \{ a \: - \: b\}} \\ \\ \\ \sf \implies {a}^{(n \: - \: 1)} \: = \: {b}^{(n \: - \: 1) } \\ \\ \\ \sf \implies \dfrac{ {a}^{(n \: - \: 1)} }{ {b}^{(n \: - \: 1)} } \: = \: 1 \\ \\ \\ \sf \implies \left \{\dfrac{a}{b} \right \}^{(n \: - \: 1)} \: = \: 1$



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



$\underline{\textsf{On Comparison,}} \\ \\ \sf \implies n \: - \: 1 \: = \: 0 \\ \\ \\ \sf \: \: \: \therefore \: \: n \: = \: 1$