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Question Number 40238 by scientist last updated on 17/Jul/18

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jul/18

$$a+(p-1)d=\frac{1}{q}$$$$a+(q-1)d=\frac{1}{p}$$$$substructing...$$$$(p-q)d=\frac{p-q}{pq}$$$$d=\frac{1}{pq}$$$$a=\frac{1}{q}-(p-1)\frac{1}{pq}$$$$a=\frac{1}{q}-\frac{1}{q}+\frac{1}{pq}$$$$a=\frac{1}{pq}$$$$s=\frac{pq}{2}[2.\frac{1}{pq}+(pq-1).\frac{1}{pq}]$$$$=\frac{pq}{2}[\frac{2}{pq}+1-\frac{1}{pq}]$$$$=\frac{pq}{2}[\frac{1}{pq}+1]$$$$=\frac{1}{2}[1+pq]$$

Answered by Rio Mike last updated on 17/Jul/18

$$d=T_2-T_1$$$$=\frac{1}{pq}since(p-q)d\stackrel{}{}=\frac{p-q}{pq}$$$$and$$$$a=\frac{1}{q}-\frac{1}{q}+\frac{1}{pq}$$$$a=\frac{1}{pq}$$$$\Rightarrow \frac{pq}{2}[\frac{1}{q}+1]=\frac{1}{2}(1+pq)QED$$

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