Question

In an H.P. $T_p=q(p+q),T_q=p(p+q)$, then $p$ and $q$ are the roots of

• A. $x^2-T_{p+q}x+T_{pq}=0$
• B. $x^2-T_{p+q}x+T_{p+q}=0$
• C. $x^2-2T_{p+q}x+T_{pq}=0$
• D. $x^2-T_{pq}x+2T_{p+q}=0$

Answer 1

Answer: (B)

$x^2-T_{p+q}x+T_{p+q}=0$

$\frac{1}{T_p}=\frac{1}{q(p+q)}=A+(p-1)D$ ..........(1)
$\dfrac{1}{T_q}=\dfrac{1}{p(p+q)}=A+(q-1)D .......(2)
Solving the above, A=$\dfrac{1}{pq(p+q)}=D ........(3)
$\therefore \frac{1}{T_{p+q}}=A+(P+q-1)D=\frac{p+q}{pq(p+q)}=\frac{1}{pq}$ by (3)
$\frac{1}{T_{pq}}=A+(pq-1)D=A(1+pq-1)=\frac{1}{p+q}$ by (3)
$\therefore p+q=T_{p+q}$ and $pq=T_{p+q}$
Hence p and q are the roots of $x^2-xS+p=0$
or $x^2-x\left(T_{pq}\right)+T_{p+q}=0$