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=(1/q) a+(q−1)d=(1/p) substructing... (p−q)d=((p−q)/(pq)) d=(1/(pq)) a=(1/q)−(p−1)(1/(pq)) a=(1/q)−(1/q)+(1/(pq)) a=(1/(pq)) s=((pq)/2)[2.(1/(pq))+(pq−1).(1/(pq))] =((pq)/2)[(2/(pq))+1−(1/(pq))] =((pq)/2)[(1/(pq))+1] =(1/2)[1+pq]](https://www.tinkutara.com/question/Q40240.png)
$${a}+\left({p}−\mathrm{1}\right){d}=\frac{\mathrm{1}}{{q}} \\ $$$${a}+\left({q}−\mathrm{1}\right){d}=\frac{\mathrm{1}}{{p}} \\ $$$${substructing}… \\ $$$$\left({p}−{q}\right){d}=\frac{{p}−{q}}{{pq}} \\ $$$${d}=\frac{\mathrm{1}}{{pq}} \\ $$$${a}=\frac{\mathrm{1}}{{q}}−\left({p}−\mathrm{1}\right)\frac{\mathrm{1}}{{pq}} \\ $$$${a}=\frac{\mathrm{1}}{{q}}−\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{pq}} \\ $$$${a}=\frac{\mathrm{1}}{{pq}} \\ $$$${s}=\frac{{pq}}{\mathrm{2}}\left[\mathrm{2}.\frac{\mathrm{1}}{{pq}}+\left({pq}−\mathrm{1}\right).\frac{\mathrm{1}}{{pq}}\right] \\ $$$$\:=\frac{{pq}}{\mathrm{2}}\left[\frac{\mathrm{2}}{{pq}}+\mathrm{1}−\frac{\mathrm{1}}{{pq}}\right] \\ $$$$=\frac{{pq}}{\mathrm{2}}\left[\frac{\mathrm{1}}{{pq}}+\mathrm{1}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+{pq}\right] \\ $$
Answered by Rio Mike last updated on 17/Jul/18
![d = T_(2 ) − T_1 = (1/(pq)) since (p−q)d ^ = ((p−q)/(pq)) and a= (1/q) − (1/q) + (1/(pq)) a= (1/(pq)) ⇒ ((pq)/2)[(1/q) + 1] = (1/2)(1 + pq)QED](https://www.tinkutara.com/question/Q40241.png)
$${d}\:=\:{T}_{\mathrm{2}\:} −\:{T}_{\mathrm{1}} \\ $$$$=\:\frac{\mathrm{1}}{{pq}}\:{since}\:\left({p}−{q}\right){d}\overset{} {\:}=\:\frac{{p}−{q}}{{pq}} \\ $$$${and}\: \\ $$$${a}=\:\frac{\mathrm{1}}{{q}}\:−\:\frac{\mathrm{1}}{{q}}\:+\:\frac{\mathrm{1}}{{pq}} \\ $$$${a}=\:\frac{\mathrm{1}}{{pq}} \\ $$$$\Rightarrow\:\frac{{pq}}{\mathrm{2}}\left[\frac{\mathrm{1}}{{q}}\:+\:\mathrm{1}\right]\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\:+\:{pq}\right){QED} \\ $$