The p^th term T_p of HP is q(p + q) and q^th term T_q is p (p + q) when p > 1, q > 1, (p ≠ q) then

T_p of AP $=\frac{1}{\mathrm{q}(\mathrm{p}+\mathrm{q})}=\mathrm{A}+(\mathrm{p}-1)\mathrm{D}$
${\text{T}}_{\mathrm{q}} \mathrm{of} \mathrm{AP}=\frac{1}{\mathrm{P}(\mathrm{p}+\mathrm{q})}=\mathrm{A}+(\mathrm{q}-1)\mathrm{D}$
solving Eqs. we get $\mathrm{A}=\mathrm{D}=\frac{1}{\mathrm{pq}(\mathrm{p}+\mathrm{q})}$
$\begin{array}{l}\frac{1}{{\mathrm{T}}_{\mathrm{p}+\mathrm{q}}}=\mathrm{A}+(\mathrm{p}+\mathrm{q}-1)\mathrm{D}\cdot \frac{1}{{\mathrm{T}}_{\mathrm{pq}}}=\mathrm{A}+(\mathrm{pq}-1)\mathrm{D}\\ \therefore \frac{1}{{\mathrm{T}}_{\mathrm{p}+\mathrm{q}}}=\mathrm{A}+(\mathrm{p}+\mathrm{q}-1)\mathrm{D}=(\mathrm{p}+\mathrm{q})\mathrm{D}=\frac{1}{\mathrm{pq}}\\ \mathrm{and} \frac{1}{{\mathrm{T}}_{\mathrm{pq}}}=\mathrm{A}+\mathrm{f}(\mathrm{p}+\mathrm{q}-1)\mathrm{D}=\mathrm{pqD}=\frac{1}{\mathrm{p}+\mathrm{q}}\\ \Rightarrow {\mathrm{T}}_{\mathrm{p}+\mathrm{q}}=\mathrm{pq} \mathrm{and} {\mathrm{T}}_{\mathrm{pq}}=\mathrm{p}+\mathrm{q} \mathrm{Also} \\ \because \mathrm{pq}>\mathrm{p}+\mathrm{q} \mathrm{i}.\mathrm{e}, {\mathrm{T}}_{\mathrm{p}+\mathrm{q}}>{\mathrm{T}}_{\mathrm{pq}}\end{array}$