Question

Find n so that $\dfrac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}}$ may be the arithmetic mean between ‘a’ and ‘b’.

Answer 1

Hint: Arithmetic mean of the two numbers is always lie in between two numbers. It can be obtained by dividing sum of given terms by given number of terms. For given problem we use the arithmetic mean formula for two numbers and then simplifying equations using laws of exponents to get required value of n.
Formulas Used: Arithmetic Mean of two numbers ‘a’ and ‘b’ is given by$\dfrac{{a + b}}{2}$
In exponents, if ${x^a}.{x^b} = {x^{a + b}}$ and in exponents if ${x^m} = {x^n}$ then$m = n$.

Complete step-by-step solution:
Let $\dfrac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}}$ be the arithmetic mean of two numbers ‘a’ and ‘b’.
Then, by definition we have$A.M. = \dfrac{{a + b}}{2}$ , substituting value of AM we have
 $\dfrac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}}$ = $\dfrac{{a + b}}{2}$ cross multiplying both side

$2\left( {{a^n} + {b^n}} \right) = \left( {a + b} \right)\left( {{a^{n - 1}} + {b^{n - 1}}} \right)$ Simplifying the brackets on both sides
$ \Rightarrow 2{a^n} + 2{b^n} = a\left( {{a^{n - 1}} + {b^{n - 1}}} \right) + b\left( {{a^{n - 1}} + {b^{n - 1}}} \right)$
$ \Rightarrow 2{a^n} + 2{b^n} = {a^{1 + n - 1}} + a.{b^{n - 1}} + b.{a^{n - 1}} + {b^{1 + n - 1}}$
$ \Rightarrow 2{a^n} + 2{b^n} = {a^n} + a.{b^{n - 1}} + b.{a^{n - 1}} + {b^n}$ (Shifting ${a^n}$and ${b^n}$ on left hand side)

$2{a^n} + 2{b^n} - {a^n} - {b^n} = a.{b^{n - 1}} + b.{a^{n - 1}}$
$ \Rightarrow {a^n} + {b^n} = a.{b^{n - 1}} + b.{a^{n - 1}}$, shifting terms having ${a^n}$ on one sides and terms having ${b^n}$ on other sides.

${a^n} - b.{a^{n - 1}} = a.{b^{n - 1}} - {b^n}$ , or we can write it as
${a^{n - 1}}.a - b.{a^{n - 1}} = a.{b^{n - 1}} - b.{b^{n - 1}}$, taking common ${a^{n - 1}}$ from left side and ${b^{n - 1}}$ from right side of the equation

${a^{n - 1}}\left( {a - b} \right) = {b^{n - 1}}\left( {a - b} \right)$
$ \Rightarrow {a^{n - 1}} = \dfrac{{{b^{n - 1}}\left( {a - b} \right)}}{{(a - b)}}$,
$ \Rightarrow {a^{n - 1}} = {b^{n - 1}}$, shifting ${b^{n - 1}}$ on left hand side we have

$\dfrac{{{a^{n - 1}}}}{{{b^{n - 1}}}} = 1$
Writing $1$as ${\left( {\dfrac{a}{b}} \right)^0}$ (we know that value of any exponent having power zero is always 1)
$ \Rightarrow \dfrac{{{a^{n - 1}}}}{{{b^{n - 1}}}} = {\left( {\dfrac{a}{b}} \right)^0}$
$ \Rightarrow {\left( {\dfrac{a}{b}} \right)^{n - 1}} = {\left( {\dfrac{a}{b}} \right)^0}$

Using law of exponent if ${a^m} = {a^n}$ then m = n we have
$n - 1 = 0$
$ \Rightarrow n = 1$
Hence, if $\dfrac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}}$ be the arithmetic mean of the two numbers then the value of n will be$1$.

Note: While, using laws of exponents one should apply laws of exponents very carefully. As we know that exponents under multiplication or division powers get either added or subtracted for terms having the same base. Also, two exponents which are equal have equal power if their bases are the same.