[Math] If an AP, a GP and a HP have the same first term and same $(2n+1)$th terms and their $n$th terms are $a,b,c$ respectively

Question

If an arithmetic preogression, a geometric progression and a harmonic preogression have the same first term and same $(2n+1)$th terms and their $n$th terms are $a,b,c$ respectively, then find the radius of the circle $x^2+y^2+2bx+2ky +ac=0$

Let the first term of the given sequences is X; therefore as per the problem

$T_1 =X$

$(2n+1)$th term of AP is $X +(2n)d $ (where $d$ is common difference)

$(2n+1)$th term of GP is $Xr^{2n} $ (where $r$ is common ratio of GP)

and $(2n+1)$th term of HP is $\frac{1}{X+(2n)D}$ where $D$ is common difference of harmonic progression.

Now $n$th term of AP is $X+(n-1)d =a ….(i)$ ;

nth term of GP is $Xr^{n-1} = b ….(ii) $

nth term of HP is $\frac{1}{X+(n-1)D} =c……(iii)$

Now the given equation which is $x^2+y^2+2bx+2ky +ac=0$ can be written as :

$(x+b)^2+(y+k)^2 = b^2+k^2-ac$

where right hand side of this equation is radius of the circle

Now how can we proceed further please suggest ………thanks.

Answer 1

Hint: if $2n+1$ terms are all equal to $A$ then $$X+2nd=Xr^{2n}=1/(X+2nD)=A$$ from which you get $d=(A-X)/2n$, $r=(A/X)^{1/2n}$, and $D={((1/A) -X})/2n$.

Now if $a,b,c$ are the $n$-th terms, i.e. just the one just before the middle term then you can calculate the middle term and back up just one term. For example the middle term of the arithmetic progression will be $(X+A)/2$, and for the geometric progression it will be $\sqrt {XA}$, Now you find the one for the harmonic case and see.