If a, b and c form a harmonic progressions what is the value of $\frac{3a+2b}{2a-b} + \frac{3c+2b}{2c-b}$?
I tried to do this by substituting $a = \frac{1}{p-q}$, $b=\frac{1}{p}$ and $c=\frac{1}{p+q}$. I got the result $\frac{5p-2q}{p+q} + \frac{5p+2q}{p-q}$. According to my solution, I am supposed to get $10 + \frac{14q^2}{p^2-q^2}$. What should I do?
Your result is correct.
Notice that $\frac{5p-2q}{p+q}+\frac{5p+2q}{p-q}$ can be written as
$$=\frac{5p^2-2pq-5pq+2q^2+5p^2+2pq+5pq+2q^2}{p^2-q^2}$$ $$=\frac{5p^2+5p^2+2q^2+2q^2-2pq+2pq-5pq+5pq}{p^2-q^2}$$ $$=\frac{10p^2+4q^2}{p^2-q^2}$$ which is nothing but $$10+\frac{14q^2}{p^2-q^2}$$