Question

If 2010 is a root of ${x^2}\left( {1 - pq} \right) - x\left( {{p^2} + {q^2}} \right) - \left( {1 + pq} \right) = 0$ and 2010 harmonic means are inserted between $p$ and $q$ the value of $\dfrac{{h_1 - h_n}}{{pq\left( {p - q} \right)}}$ isA) $\dfrac{1}{2}$B) 4C) 1D) 2

Hint: First we will let the given root be equal to n then we will replace $x$ by $n$ in the given equation and get a certain value. Now since 2010 harmonic means are inserted between $p$ and $q$ we will assume them as $h_1,h_2,....h_n$ and then convert them in A.P. Then use certain formulas of A.P to get the desired value.
The identity used here is :
${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$

Complete step by step solution:
The given equation is:-
${x^2}\left( {1 - pq} \right) - x\left( {{p^2} + {q^2}} \right) - \left( {1 + pq} \right) = 0$
Let the given root to be $n$
Therefore,
$n = 2010$
Now since n is a root of given equation therefore replacing $x$ by $n$ in the given equation we get:
${n^2}\left( {1 - pq} \right) - n\left( {{p^2} + {q^2}} \right) - \left( {1 + pq} \right) = 0$
Solving it further we get:
$\Rightarrow {n^2} - {n^2}pq - n{p^2} - n{q^2} - 1 - pq = 0 \\ \Rightarrow {n^2} - 1 = pq\left( {{n^2} + 1} \right) + n\left( {{p^2} + {q^2}} \right).................\left( 1 \right) \\$
Now since 2010 harmonic means are inserted between p and q therefore let the harmonic means be:
$h_1,h_2,...........h_n$
Hence the resulting harmonic series is:
$p,h_1,h_2,...........h_n,q$
Now converting this series into A.P. we get:-
$\Rightarrow \dfrac{1}{p},\dfrac{1}{{h_1}},\dfrac{1}{{h_2}},...........\dfrac{1}{{h_n}},\dfrac{1}{q}$
Now as we know that the formula for last term of an A.P is :-
$Tn = a + \left( {N - 1} \right)d$
Applying this formula for above A.P we get:
Here, $N = n + 2$ hence
$\Rightarrow \dfrac{1}{q} = \dfrac{1}{p} + \left( {n + 2 - 1} \right)d \\ \Rightarrow \dfrac{1}{q} = \dfrac{1}{p} + \left( {n + 1} \right)d \\$
Solving for the value of d we get:-
$\Rightarrow \left( {n + 1} \right)d = \dfrac{1}{q} - \dfrac{1}{p} \\ \Rightarrow \left( {n + 1} \right)d = \dfrac{{p - q}}{{pq}} \\ \Rightarrow d = \dfrac{{p - q}}{{pq\left( {n + 1} \right)}} \\$
Now since is the second term of the above A.P therefore,
$\Rightarrow \dfrac{1}{{h_1}} = \dfrac{1}{p} + \left( {2 - 1} \right)d \\ \Rightarrow \dfrac{1}{{h_1}} = \dfrac{1}{p} + d \\$
Putting the value of d we get:-
$\Rightarrow \dfrac{1}{{h_1}} = \dfrac{1}{p} + \dfrac{{p - q}}{{pq\left( {n + 1} \right)}} \\ \Rightarrow \dfrac{1}{{h_1}} = \dfrac{{q\left( {n + 1} \right) + p - q}}{{pq\left( {n + 1} \right)}} \\ \Rightarrow \dfrac{1}{{h_1}} = \dfrac{{qn + p}}{{pq\left( {n + 1} \right)}} \\ \Rightarrow h_1 = \dfrac{{pq\left( {n + 1} \right)}}{{qn + p}} \\$
Also, since $\dfrac{1}{{h_n}}$ is the second last term of the above A.P therefore,
$\Rightarrow \dfrac{1}{{h_n}} = \dfrac{1}{q} - d$
Putting the value of d we get:-
$\Rightarrow \dfrac{1}{{h_n}} = \dfrac{1}{q} - \dfrac{{p - q}}{{pq\left( {n + 1} \right)}} \\ \Rightarrow \dfrac{1}{{h_n}} = \dfrac{{p\left( {n + 1} \right) - p + q}}{{pq\left( {n + 1} \right)}} \\ \Rightarrow \dfrac{1}{{h_n}} = \dfrac{{pn + q}}{{pq\left( {n + 1} \right)}} \\ \Rightarrow h_n = \dfrac{{pq\left( {n + 1} \right)}}{{pn + q}} \\$
Now evaluating the value of $h_1 - h_n$ we get:-
$\Rightarrow h_1 - h_n = \dfrac{{pq\left( {n + 1} \right)}}{{qn + p}} - \dfrac{{pq\left( {n + 1} \right)}}{{pn + q}} \\ \Rightarrow h_1 - h_n = \dfrac{{pq\left( {n + 1} \right)\left( {pn + q - qn - p} \right)}}{{\left( {qn + p} \right)\left( {pn + q} \right)}} \\ \Rightarrow h_1 - h_n = \dfrac{{pq\left( {n + 1} \right)\left( {n\left( {p - q} \right) - 1\left( {p - q} \right)} \right)}}{{\left( {qn + p} \right)\left( {pn + q} \right)}} \\ \Rightarrow h_1 - h_n = \dfrac{{pq\left( {n + 1} \right)\left( {n - 1} \right)\left( {p - q} \right)}}{{{n^2}pq + n{q^2} + n{p^2} + pq}} \\$
As we know that:
$\Rightarrow {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
Therefore applying this formula we get:
$\Rightarrow h_1 - h_n = \dfrac{{pq\left( {{n^2} - 1} \right)\left( {p - q} \right)}}{{({n^2} + 1)pq + n({q^2} + {p^2})}}$
Now putting the value of $\left( {{n^2} - 1} \right)$ from equation 1 we get:-
$\Rightarrow h_1 - h_n = \dfrac{{pq\left[ {pq\left( {{n^2} + 1} \right) + n\left( {{p^2} + {q^2}} \right)} \right]\left( {p - q} \right)}}{{({n^2} + 1)pq + n({q^2} + {p^2})}}$
Now evaluating the value of $\dfrac{{h_1 - h_n}}{{pq\left( {p - q} \right)}}$ we get:-
$\Rightarrow \dfrac{{h_1 - h_n}}{{pq\left( {p - q} \right)}} = \dfrac{{pq\left[ {pq\left( {{n^2} + 1} \right) + n\left( {{p^2} + {q^2}} \right)} \right]\left( {p - q} \right)}}{{({n^2} + 1)pq + n({q^2} + {p^2})pq\left( {p - q} \right)}} \\ \Rightarrow \dfrac{{h_1 - h_n}}{{pq\left( {p - q} \right)}} = \dfrac{{\left[ {pq\left( {{n^2} + 1} \right) + n\left( {{p^2} + {q^2}} \right)} \right]}}{{({n^2} + 1)pq + n({q^2} + {p^2})}} \\ \Rightarrow \dfrac{{h_1 - h_n}}{{pq\left( {p - q} \right)}} = 1 \\$

$\therefore$ Hence the option C is correct.

Note:
Harmonic terms are the reciprocal of the terms that are in A.P.
Also, we do not have to use the exact value of the root of the given equation to get the desired answer.
The nth term of an A.P. is given by:
$Tn = a + \left( {N - 1} \right)d$ where N is the number of terms of A.P.