# Problem in harmonic series.

If the sum of roots of the quadratic equation $ax^2 +bx+c=0$

{ $(a,b,c)$ not equal to zero}

is equal to sum of squares of their reciprocals , then $a/c$ , $b/a$, $c/b$ are in?

Actually the options given were -

$a.p$ ,

$g.p$ ,

$h.p$

none of these.

I could make it out that $a.p$ and $g.p$ are not the answers.

I know a little about harmonic series that it's like the reciprocal of an arithmetic series but i don't know much and hence couldn't solve it.

Any help is appreciated.

• Maybe you can try to formulate your question in a more understandable English? Sorry but I didn't really get your question.... – Umberto Oct 12 '13 at 8:25

$a \over c$, $b\over a$,$c\over b$ are in HP means that $c\over a$,$a\over b$,$b\over c$ are in AP.This should simplify things.
Let the two roots be $l$ and $m$ then:
$$l+m={-b\over a}$$ $$l.m={c\over a}$$ $$l^2+m^2={{b^2\over a^2}-2*{c\over a}}$$ If sum of roots is equal to sum of squares of their reciprocals then: $$l+m={{1\over l^2}+{1\over m^2}}$$ Solve this using the above 3 equations to get a relation between $a/c,b/a,c/b$.