Question Number 41966 by Raj Singh last updated on 16/Aug/18
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
![S_p =(p/2)[2a+(p−1)d] S_Q =(Q/2)[2a+(Q−1)d] given S_p =S_Q (p/2)[2a+(p−1)d]=(Q/2)[2a+(Q−1)d] 2pa+(p^2 −p)d−2Qa−(Q^2 −Q)d=0 2a(P−Q)+d(P^2 −P−Q^2 +Q)=0 2a(P−Q)+d{(P+Q)(P−Q)−(P−Q)}=0 2a(P−Q)+(P−Q){P+Q−1}d=0 2a+(P+Q−1)d=0 nos S_(P+Q) =((P+Q)/2)[2a+(P+Q−1)d] =((P+Q)/2)×0=0](https://www.tinkutara.com/question/Q41967.png)
$${S}_{{p}} =\frac{{p}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({p}−\mathrm{1}\right){d}\right] \\ $$$${S}_{{Q}} =\frac{{Q}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({Q}−\mathrm{1}\right){d}\right] \\ $$$${given}\:\:{S}_{{p}} ={S}_{{Q}} \\ $$$$\frac{{p}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({p}−\mathrm{1}\right){d}\right]=\frac{{Q}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({Q}−\mathrm{1}\right){d}\right] \\ $$$$\mathrm{2}{pa}+\left({p}^{\mathrm{2}} −{p}\right){d}−\mathrm{2}{Qa}−\left({Q}^{\mathrm{2}} −{Q}\right){d}=\mathrm{0} \\ $$$$\mathrm{2}{a}\left({P}−{Q}\right)+{d}\left({P}^{\mathrm{2}} −{P}−{Q}^{\mathrm{2}} +{Q}\right)=\mathrm{0} \\ $$$$\mathrm{2}{a}\left({P}−{Q}\right)+{d}\left\{\left({P}+{Q}\right)\left({P}−{Q}\right)−\left({P}−{Q}\right)\right\}=\mathrm{0} \\ $$$$\mathrm{2}{a}\left({P}−{Q}\right)+\left({P}−{Q}\right)\left\{{P}+{Q}−\mathrm{1}\right\}{d}=\mathrm{0} \\ $$$$\mathrm{2}{a}+\left({P}+{Q}−\mathrm{1}\right){d}=\mathrm{0} \\ $$$${nos}\:{S}_{{P}+{Q}} =\frac{{P}+{Q}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({P}+{Q}−\mathrm{1}\right){d}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{P}+{Q}}{\mathrm{2}}×\mathrm{0}=\mathrm{0} \\ $$