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If an A.P , a G.P and a H.P have the same first term and same (2n+1)th terms and their nth 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 A.P. = $X +(2n)d $ ( where d is common difference)

$(2n+1)th $ term of G.P. = $Xr^{2n} $ ( where r is common ratio of G.P. )

and (2n+1)th term of H.P. = $\frac{1}{X+(2n)D}$ where D is common difference of Harmonic progression.

Now nth term of A.P. = $X+(n-1)d =a ....(i)$ ;

nth term of G.P. = $Xr^{n-1} = b ....(ii) $

nth term of H.P. = $\frac{1}{X+(n-1)D} =c......(iii)$

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 R.H.S 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

1 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.

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