Given ap={tex}1\\over q{/tex} aq={tex}1\\over p{/tex}\xa0let first term and common difference of given AP is a and d respectively.Then ap={tex}1\\over q{/tex}\xa0 a+(p-1)d={tex}1\\over q{/tex}\xa0——–(1)and aq=a+(q-1)d={tex}1\\over p{/tex} ——–(2)On subtracting eq(2) from (1) we geta+(p-1)d -(a+(q-1)d) ={tex}1\\over q{/tex}\xa0{tex}-{1\\over p}{/tex}a+pd-d-a-qd+d={tex}p-q\\over pq{/tex}pd-qd={tex}p-q\\over pq{/tex}d(p-q)={tex}p-q\\over pq{/tex} d={tex}1\\over pq{/tex}——–(3)Substituting value of d in eq(1) we geta+(p-1)×{tex}1\\over pq{/tex}\xa0={tex}1\\over q{/tex}a+p×{tex}1\\over pq{/tex}\xa0-{tex}1\\over pq{/tex}\xa0={tex}1\\over q{/tex}\xa0a+{tex}1\\over q{/tex}\xa0-{tex}1\\over pq{/tex}=\xa0{tex}1\\over q{/tex} a={tex}{1\\over q}-{1\\over q}+{1\\over pq}{/tex} a={tex}1\\over pq{/tex} ——(4)Now Spq={tex}pq\\over 2{/tex}\xa0[2a+(pq-1)d] ={tex}pq\\over 2{/tex}\xa0[2×{tex}1\\over pq{/tex}\xa0+(pq-1)×{tex}1\\over pq{/tex}\xa0] =\xa0{tex}pq\\over 2{/tex}[{tex}{2+(pq-1)\\over pq}{/tex}] ={tex}{pq×(pq+1)\\over 2×pq}{/tex} ={tex}pq+1\\over 2{/tex} H.Proved
P+q/2pq