Question

Can anyone please answer ??

Answer 1

$\Large\underline\mathfrak{Question}$

$\frac{ {a}^{n + 1} + {b}^{n + 1} }{ {a}^{n} + {b}^{n} } \: is \: AM \: of \: a \: and \: b. \\ \\ find \: n.$

$\Large\bold\star\underline{\underline\textbf{Formula\:used}}$

• a^(x+y) = a^x * a^y
• Arithmetic Mean of 2 numbers a and b is (a+b)/2 .
• a^0 = 1.

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

Using a^(x+y) = a^x * a^y in numerator of Question,

and, than putting them Equal to (a+b)/2 we get,,

$\frac{ {a}^{n + 1} + {b}^{n + 1} }{ {a}^{n} + {b}^{n} } = \frac{a + b}{2} \\ \\ \red\leadsto \: \frac{ {a}^{n}a + {b}^{n } b }{ {a}^{n} + {b}^{n} } \: = \frac{a + b}{2} \\ \\ \green{\textbf{cross - multiply \: now}} \\ \\ \red\leadsto \: 2 ({a}^{n}a + {b}^{n } b ) = {a}^{n}a + {a}^{n}b + {b}^{n}b + {b}^{n}a \\ \\ \red\leadsto \: {a}^{n}a + {b}^{n } b = {a}^{n}b + {b}^{n}a \\ \\ \red\leadsto \: {a}^{n}a - {a}^{n}b = {b}^{n}a - {b}^{n } b \\ \\ \red\leadsto \: {a}^{n} \cancel{(a - b)} = {b}^{n} \cancel{(a - b)} \\ \\ \red\leadsto \: {a}^{n} = {b}^{n} \\ \\ \red{\textbf{a and b can only be = if n=0}} \\ as \: \: {x}^{0} = 1 \\ \\ so \\ \\ \pink{\large\boxed{\boxed{\bold{n = 0}}}}$

Hence, value of n is Equal to 0.

$\large\underline\textbf{Hope it Helps You.}$