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……
Best Answer
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.