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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.

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Also, the circle equation should be $(x+b)^2+(y+k)^2 = b^2+k^2-ac$. I'll fix them for you – Pratyush Sarkar Aug 5 '13 at 15:21
I am guessing that the problem wanted to say $n+1$-st term is $a,b,c$ not the $n$-th. Then you could get an easy connection between 1,2n+1 and the middle term which is the one at n+1. – Maesumi Aug 5 '13 at 15:22
Where does $k$ come from? – Kunnysan Aug 5 '13 at 16:22
if $a, b, c$ are given, then the radius is $r=\sqrt{b^2+k^2-ac}$ is already a solution in terms of known values (apart from $k$, which, per the question above, it is unclear where it originates)... is there a point which is being missed here? – hypergeometric Aug 28 '14 at 11:58

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.

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re my comment to the OQ above, if we already know $a, b, c$ which are given, and can express the required answer using these, then why do we even need to know anything about the $(2n+1)$-th terms?It looks like an interesting question nevertheless. – hypergeometric Aug 28 '14 at 12:50
@hypergeometric I guess the statement of problems needs to be made more precise. Find $R$ in terms of ... etc. Or perhaps it is left to imagination of solver. – Maesumi Aug 29 '14 at 21:16
@@Maesumi see my solution above... probably either your suggestion or the other possibility is correct... then the problem reduces quite conveniently to a neat solution. – hypergeometric Aug 30 '14 at 5:22

If we assume that

  • either Maesumi's comment to the OQ is correct, i.e. $n$-th term should read $(n+1)$-th term
  • or that the $(2n+1)$-th term should read $(2n-1)$-th term,

then the number of terms between the three terms are equally spaced apart in terms of number of terms (no pun intended... :)), in which case the symmetry can be exploited to simplify the solution.

Let $A$ = first term; $B$ = last term; $\lbrace a,b,c \rbrace$= middle term for $\lbrace \text{AP, GP, HP}\rbrace$ respectively as given.

Then: $$\begin{align} &\text{AP}: &&a-A=B-a &\Rightarrow a&=\frac {A+B}2\\ &\text{GP}: &&\frac bA=\frac Bb &\Rightarrow b&=\sqrt{AB}\\ &\text{HP}: &&\frac 1c-\frac1A=\frac1B-\frac1c\\ & &&\frac 2c=\frac 1A-\frac 1B &\Rightarrow c&=\frac{2AB}{A+B}=\frac {b^2}{a}\Rightarrow b^2=ac\end{align}$$

Hence radius of circle, $$r=\sqrt{b^2-ac+k^2}=k\qquad \blacksquare$$

NB: Now it becomes clear what $k$ represents and why it is there in the first place.

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