Question

Question:If
$\frac{{a}^{n + 1} +b {}^{n + 1} } { {a}^{n} + {b}^{n} }$
is the A.M.between a and b. Then, find the value of n. ​

Answer 1

$\huge{ \underline{ \purple{ \bold{ \underline{ \mathrm{AnS{\blue{weR }}}}}}}} = >$

we have,

• A. M. between a and b=
• $\frac{a + b}{2}$

It is given that

• $\frac{a {}^{n + 1} + b + {}^{n + 1} }{ {a}^{n} + {b}^{n} }$is the A. M. between a and b.

• $\frac{a {}^{n + 1} + b {}^{n + 1} }{ {a}^{n } + {b}^{n} } = \frac{a + b}{2}$

$= > 2( {a}^{n + 1} + {b}^{n + 1} ) = ({a}^{n} + {b}^{n} )(a + b)$

$= > 2 {a}^{n + 1} + 2 {b}^{n + 1} = {a}^{n + 1} + a {b}^{n} + {a}^{n} b + {b}^{ + 1}$

$= > {a}^{n + 1} + {b}^{n + 1} = a {b}^{n} + {a}^{n} b$

$= > {a}^{n + 1} - {a}^{n} b = a {b}^{n} - {b}^{n} - {b}^{n + 1}$

$= > {a}^{n} (a - b) = {b}^{n} (a - b)$

$= > {a}^{n} = {b}^{n}$

$= > \frac{a {}^{n} }{ {b}^{n} } = 1 = > \binom{a}{b} {}^{n} = \binom{a}{b} {}^{0} = > n = 0$

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