If $\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$ is the $A.M.$ between $a$ and $b,$ then find the value of $n$.
If $\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$ is the $A.M.$ between $a$ and $b,$ then find the value of $n$.
$A.M.$ of $a$ and $b$ $=\frac{a+b}{2}$
According to the given condition,
$\frac{a+b}{2}=\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$
$\Rightarrow(a+b)\left(a^{n-1}+b^{n-1}\right)=2\left(a^{n}+b^{n}\right)$
$\Rightarrow a^{n}+a b^{n-1}+b a^{n-1}+b^{n}=2 a^{n}+2 b^{n}$
$\Rightarrow a b^{n-1}+a^{n-1} b=a^{n}+b^{n}$
$\Rightarrow a b^{n-1}-b^{n}=a^{n}-a^{n-1} b$
$\Rightarrow b^{n-1}(a-b)=a^{n-1}(a-b)$
$\Rightarrow b^{n-1}=a^{n-1}$
$\Rightarrow\left(\frac{a}{b}\right)^{n-1}=1=\left(\frac{a}{b}\right)^{0}$
$\Rightarrow n-1=0$
$\Rightarrow n=1$
Similar Questions
Find the sum of all natural numbers lying between $100$ and $1000,$ which are multiples of $5 .$
Find the sum of all natural numbers lying between $100$ and $1000,$ which are multiples of $5 .$
If $\frac{{3 + 5 + 7 + ..........{\rm{to}}\;n\;{\rm{terms}}}}{{5 + 8 + 11 + .........{\rm{to}}\;10\;{\rm{terms}}}} = 7$, then the value of $n$ is
If $\frac{{3 + 5 + 7 + ..........{\rm{to}}\;n\;{\rm{terms}}}}{{5 + 8 + 11 + .........{\rm{to}}\;10\;{\rm{terms}}}} = 7$, then the value of $n$ is
If the sum of three numbers of a arithmetic sequence is $15$ and the sum of their squares is $83$, then the numbers are
If the sum of three numbers of a arithmetic sequence is $15$ and the sum of their squares is $83$, then the numbers are
Let $a_1, a_2 , a_3,.....$ be an $A.P$, such that $\frac{{{a_1} + {a_2} + .... + {a_p}}}{{{a_1} + {a_2} + {a_3} + ..... + {a_q}}} = \frac{{{p^3}}}{{{q^3}}};p \ne q$. Then $\frac{{{a_6}}}{{{a_{21}}}}$ is equal to
Let $a_1, a_2 , a_3,.....$ be an $A.P$, such that $\frac{{{a_1} + {a_2} + .... + {a_p}}}{{{a_1} + {a_2} + {a_3} + ..... + {a_q}}} = \frac{{{p^3}}}{{{q^3}}};p \ne q$. Then $\frac{{{a_6}}}{{{a_{21}}}}$ is equal to
- [JEE MAIN 2013]
The number of terms of the $A.P. 3,7,11,15...$ to be taken so that the sum is $406$ is
The number of terms of the $A.P. 3,7,11,15...$ to be taken so that the sum is $406$ is