Question Number 40238 by scientist last updated on 17/Jul/18
Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jul/18
$${a}+\left({p}−\mathrm{1}\right){d}=\frac{\mathrm{1}}{{q}} \\ $$$${a}+\left({q}−\mathrm{1}\right){d}=\frac{\mathrm{1}}{{p}} \\ $$$${substructing}… \\ $$$$\left({p}−{q}\right){d}=\frac{{p}−{q}}{{pq}} \\ $$$${d}=\frac{\mathrm{1}}{{pq}} \\ $$$${a}=\frac{\mathrm{1}}{{q}}−\left({p}−\mathrm{1}\right)\frac{\mathrm{1}}{{pq}} \\ $$$${a}=\frac{\mathrm{1}}{{q}}−\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{pq}} \\ $$$${a}=\frac{\mathrm{1}}{{pq}} \\ $$$${s}=\frac{{pq}}{\mathrm{2}}\left[\mathrm{2}.\frac{\mathrm{1}}{{pq}}+\left({pq}−\mathrm{1}\right).\frac{\mathrm{1}}{{pq}}\right] \\ $$$$\:=\frac{{pq}}{\mathrm{2}}\left[\frac{\mathrm{2}}{{pq}}+\mathrm{1}−\frac{\mathrm{1}}{{pq}}\right] \\ $$$$=\frac{{pq}}{\mathrm{2}}\left[\frac{\mathrm{1}}{{pq}}+\mathrm{1}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+{pq}\right] \\ $$
Answered by Rio Mike last updated on 17/Jul/18
$${d}\:=\:{T}_{\mathrm{2}\:} −\:{T}_{\mathrm{1}} \\ $$$$=\:\frac{\mathrm{1}}{{pq}}\:{since}\:\left({p}−{q}\right){d}\overset{} {\:}=\:\frac{{p}−{q}}{{pq}} \\ $$$${and}\: \\ $$$${a}=\:\frac{\mathrm{1}}{{q}}\:−\:\frac{\mathrm{1}}{{q}}\:+\:\frac{\mathrm{1}}{{pq}} \\ $$$${a}=\:\frac{\mathrm{1}}{{pq}} \\ $$$$\Rightarrow\:\frac{{pq}}{\mathrm{2}}\left[\frac{\mathrm{1}}{{q}}\:+\:\mathrm{1}\right]\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\:+\:{pq}\right){QED} \\ $$