Find n, so that `(a^(n+1)+b^(n+1))/(a^(n)+b^(n))(a ne b )` be HM beween a and b.

Explanation

` therefore (a^(n+1)+b^(n+1))/(a^(n)+b^(n))=(2ab)/(a+b)`
` implies (b^(n+1)[((a)/(b))^(n+1)+1])/(b^(n)[b^(n)((a)/(b)+1)])=((b)^(2)[2((a)/(b))])/(b((a)/(b)+1))`
` implies (((a)/(b))^(n+1)+1)/(((a)/(b))^(n)+1)=(2((a)/(b)))/(((a)/(b))+1)`
Let `(a)/(b)=lambda`
Then, ` (lambda^(n+1)+1)/(lambda^(n)+1)=(2lambda)/(lambda+1)`
`implies (lambda+1)(lambda^(n+1)+1)=(2lambda)(lambda^(n)+1)`
`implies lambda^(n+2)+lambda+lambda^(n+1)+1 =2lambda^(n+1)+2lambda`
`implies lambda^(n+2)-lambda^(n+1)-lambda+1 =0`
`implies lambda^(n+1)(lambda-1)-1(-1)=0`
`implies(lambda-1) (lambda^(n+1)-1)=0`
`implies lambda-1ne 0 " " [therefore ane b]`
`therefore lambda^(n+1)-1=0`
`implies lambda^(n+1)=1=lambda^(0)`
`implies n+1=0" or "n=-1`