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If an A.P ( Arithmetic Progression) , a G.P( Geometric Progression) and a H.P ( Harmonic Progression) have the same first term and same (2n+1)th terms and their nth terms are a,b,c respectively , then the radius of the circle $x^2+y^2+2bx+2ky +ac=0$ is ( options are given below)

(a) k

(b) $\boldsymbol{|k|}$

(c) $\sqrt{b^2-ac}$

(d) none of these

My approach :

Writing the equation of circle in standard form viz. $(x-a)^2+(y-b)^2 =r^2$ where a,b are centre of the circle and r is its radius.

we have $(x-b)^2+(y-k)^2 = -ac +b^2+k^2$........(i)

Also let the first term of A.P , G.P. & H.P is x ( as they have same first term) and let z be the (2n+1)th term of A.P. G.P. \& H.P.

Now, (2n+1)th term of A.P. $\Rightarrow x +(2n)d =z .....(ii)$

(2n+1)th term of G.P $\Rightarrow xr^{2n} =z .....(iii)$

(2n+1)th term of H.P. $\Rightarrow \frac{1}{x+2nd} =z.......(iv)$

Now nth term of A.P. $\Rightarrow x +(n-1)d = a .......(v)$( where d is common difference) ;

nth term of G.P. $\Rightarrow xr^{n-1} = b........(vi) $ ( where r is the common ratio)

nth term of H.P $\Rightarrow \frac{1}{x+(n-1)d} =c..........(vii) $

Multiplying $(ii) & (iv)$ we get $z^2 = 1 \Rightarrow z = \pm 1......(viii)$

Also multiplying $(v) & (vii) $ we get ac = 1.$........(ix) $

Now what to do next ......please suggest further... thanks......

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I suggest you find $n$, $d$ and $r$ from system of ($ii$), ($iii$) and ($iv$). –  Juris Jul 31 '13 at 11:21
    
It's not a very good idea to use the same letter to denote possibly different things...$\,x,$ as variable and as the first term in the progressions, say. Also the $\,a\,$ as first coordinate of the circle's center and as the $\,n-$th term of the AP...this is going to get pretty messed up. –  DonAntonio Jul 31 '13 at 11:26
    
@DonAntonio, and also r :) –  Juris Jul 31 '13 at 11:52
    
An A.P., a G.P., and an H.P. walk into a bar ... –  marty cohen Nov 24 '13 at 21:58

2 Answers 2

Hint: With $k$ arbitrary your answer must depend on $k$, and using the equation of the circle, it is easy to formulate a condition for the answer to depend on $k$ alone. It is then a question of checking whether that condition applies.

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First off, I suspect you misstated the question and mean for $a$, $b$, and $c$ to be the $n+1$st (middle) terms of the sequences. As you stated the question, a single non-trivial example will quickly exclude answers a), b), and c).

Next, look at a simple example. The sequence (1,1,1), or even the sequence (1), is at once arithmetic, geometric, and harmonic, and therefore $a=b=c=1$ is one possibility. The circle $x^2+y^2+2x+2ky+1=0$ has standard equation $(x+1)^2+(x-k)^2=k^2$, so the radius is $|k|$. For $k=-1$, the radius is neither $\sqrt{b^2-ac}=0$ nor $k$, so the correct answer is either b) or d).

To rule out d) requires a proof, since it could be the case that the radius is often, but not always, $|k|$.

Worth observing is that for a sequence that's harmonic/geometric/arithmetic, the three-term sequence comprising the 1st, $n$-th, and $2n+1$st term is also harmonic/geometric/arithmetic. If $s$ and $t$ are the common first and last terms of the sequences, $a=\displaystyle\frac{s+t}{2}$, $b=\sqrt{st}$, and $c=\displaystyle\frac{2st}{s+t}$. Perhaps you can take it from there.

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