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Question

In an H.P. Tp=q(p+q),Tq=p(p+q), then p and q are the roots of

A
x2Tp+qx+Tpq=0
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B
x2Tp+qx+Tp+q=0
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C
x22Tp+qx+Tpq=0
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D
x2Tpqx+2Tp+q=0
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Solution

The correct option is A x2Tp+qx+Tp+q=0
1Tp=1q(p+q)=A+(p1)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)
1Tp+q=A+(P+q1)D=p+qpq(p+q)=1pq by (3)
1Tpq=A+(pq1)D=A(1+pq1)=1p+q by (3)
p+q=Tp+q and pq=Tp+q
Hence p and q are the roots of x2xS+p=0
or x2x(Tpq)+Tp+q=0

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