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,
$\frac{a+b}{2}=\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$
$\Rightarrow(a+b)\left(a^{n-1}+b^{n-1}\right)=2\left(a^{n}+b^{n}\right)$
$\Rightarrow a^{n}+a b^{n-1}+b a^{n-1}+b^{n}=2 a^{n}+2 b^{n}$
$\Rightarrow a b^{n-1}+a^{n-1} b=a^{n}+b^{n}$
$\Rightarrow a b^{n-1}-b^{n}=a^{n}-a^{n-1} b$
$\Rightarrow b^{n-1}(a-b)=a^{n-1}(a-b)$
$\Rightarrow b^{n-1}=a^{n-1}$
$\Rightarrow\left(\frac{a}{b}\right)^{n-1}=1=\left(\frac{a}{b}\right)^{0}$
$\Rightarrow n-1=0$
$\Rightarrow n=1$
The sum of the common terms of the following three arithmetic progressions.
$3,7,11,15,...................,399$
$2,5,8,11,............,359$ and
$2,7,12,17,...........,197$, is equal to $................$.
The Fibonacci sequence is defined by
$1 = {a_1} = {a_2}{\rm{ }}$ and ${a_n} = {a_{n - 1}} + {a_{n - 2}},n\, > \,2$
Find $\frac{a_{n+1}}{a_{n}},$ for $n=1,2,3,4,5$
If the sum and product of the first three term in an $A.P$. are $33$ and $1155$, respectively, then a value of its $11^{th}$ tern is
If the numbers $a,\;b,\;c,\;d,\;e$ form an $A.P.$, then the value of $a - 4b + 6c - 4d + e$ is
If $n$ arithmetic means are inserted between a and $100$ such that the ratio of the first mean to the last mean is $1: 7$ and $a+n=33$, then the value of $n$ is