Solveeit Logo
Login

Question

Question: Find n so that \(\dfrac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}}\) may be the arithmetic mean b...

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

Explanation

Solution

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 bya+b2\dfrac{{a + b}}{2}
In exponents, if xa.xb=xa+b{x^a}.{x^b} = {x^{a + b}} and in exponents if xm=xn{x^m} = {x^n} thenm=nm = n.

Complete step-by-step solution:
Let an+bnan−1+bn−1\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 haveA.M.=a+b2A.M. = \dfrac{{a + b}}{2} , substituting value of AM we have
an+bnan−1+bn−1\dfrac{{{a^n} + {b^n}}}{{{a^{n - 1}} + {b^{n - 1}}}} = a+b2\dfrac{{a + b}}{2} cross multiplying both side

2(an+bn)=(a+b)(an−1+bn−1)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
⇒2an+2bn=a(an−1+bn−1)+b(an−1+bn−1)\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)
⇒2an+2bn=a1+n−1+a.bn−1+b.an−1+b1+n−1\Rightarrow 2{a^n} + 2{b^n} = {a^{1 + n - 1}} + a.{b^{n - 1}} + b.{a^{n - 1}} + {b^{1 + n - 1}}
⇒2an+2bn=an+a.bn−1+b.an−1+bn\Rightarrow 2{a^n} + 2{b^n} = {a^n} + a.{b^{n - 1}} + b.{a^{n - 1}} + {b^n} (Shifting an{a^n}and bn{b^n} on left hand side)

2an+2bn−an−bn=a.bn−1+b.an−12{a^n} + 2{b^n} - {a^n} - {b^n} = a.{b^{n - 1}} + b.{a^{n - 1}}
⇒an+bn=a.bn−1+b.an−1\Rightarrow {a^n} + {b^n} = a.{b^{n - 1}} + b.{a^{n - 1}}, shifting terms having an{a^n} on one sides and terms having bn{b^n} on other sides.

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

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

an−1bn−1=1\dfrac{{{a^{n - 1}}}}{{{b^{n - 1}}}} = 1
Writing 11as (ab)0{\left( {\dfrac{a}{b}} \right)^0} (we know that value of any exponent having power zero is always 1)
⇒an−1bn−1=(ab)0\Rightarrow \dfrac{{{a^{n - 1}}}}{{{b^{n - 1}}}} = {\left( {\dfrac{a}{b}} \right)^0}
⇒(ab)n−1=(ab)0\Rightarrow {\left( {\dfrac{a}{b}} \right)^{n - 1}} = {\left( {\dfrac{a}{b}} \right)^0}

Using law of exponent if am=an{a^m} = {a^n} then m = n we have
n−1=0n - 1 = 0
⇒n=1\Rightarrow n = 1
Hence, if an+bnan−1+bn−1\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 be11.

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.