Find the value of n so that `(a^(n+1)+b^(n+1))/(a^(n)+b^n)` may be the geometric mean between a and b.

If (a^(n + 1) + b^(n + 1))/(a^(n) + b^(n)) is the arithmetic mean between a and b , then n =

If (a^(n)+b^n)/ (a^(n-1) +b^(n-1)) is the A.M. between a and b, then find the value of n.

Find the value of n in N such that the curve (x/a)^n+(y/b)^n=2 touches the straight line x/a+y/b=2 at the point (a , b)dot

Find the value of n if (i) ( n + 1) ! = 20 ( n-1 ) ! ( ii) 1/( 8!) + 1/( 9 !) = n/( 10 !)

Let a_(1), a_(2) ...be positive real numbers in geometric progression. For n, if A_(n), G_(n), H_(n) are respectively the arithmetic mean, geometric mean and harmonic mean of a_(1), a_(2),..., a_(n) . Then, find an expression for the geometric mean of G_(1), G_(2),...,G_(n) in terms of A_(1), A_(2),...,A_(n), H_(1), H_(2),..., H_(n)

If a > b >0, with the aid of Lagranges mean value theorem, prove that n b^(n-1)(a-b) < a^n -b^n < n a^(n-1)(a-b) , if n >1. n b^(n-1)(a-b) > a^n-b^n > n a^(n-1)(a-b) , if 0 < n < 1.