Question Number 43348 by peter frank last updated on 10/Sep/18
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Sep/18
![3)2((b/a)+(b/c))=(a/b)+(a/c)+(c/a)+(c/b) p=(b/a) q=(b/c) (q/p)=(a/c) (a/q)=(c/p)=r(say) a=rq c=pr b=ap=(rq)p=pqr b=qc=q(pr)=pqr a=rq b=pqr c=pr 2((b/a)+(b/c))=(a/b)+(a/c)+(c/a)+(c/b) 2(((pqr)/(rq))+((pqr)/(pr)))=((rq)/(pqr))+((rq)/(pr))+((pr)/(rq))+((pr)/(pqr)) 2(p+q)=(1/p)+(q/p)+(p/q)+(1/q) 2(p+q)=((p+q)/(pq))+((p^2 +q^2 )/(pq)) 2pq(p+q)=p+q+p^2 +q^2 2pq(p+q)+2pq=p+q+p^2 +q^2 +2pq 2pq(p+q+1)=p+q+(p+q)^2 2pq(p+q+1)=(p+q)(1+p+q) (p+q+1)(2pq−p−q)=0 when 2pq−p−q=0 2pq=p+q [p+q+1 can not be equsls to zero] now 2pa=p+q multiply both side by r 2pqr=pr+qr 2b=c+a [since pqr=b pr=c qr=a] so b−a=c−b hence a,b,c are in AP](Q43383.png)
$$\left.\mathrm{3}\right)\mathrm{2}\left(\frac{{b}}{{a}}+\frac{{b}}{{c}}\right)=\frac{{a}}{{b}}+\frac{{a}}{{c}}+\frac{{c}}{{a}}+\frac{{c}}{{b}} \\ $$$${p}=\frac{{b}}{{a}}\:\:\:\:{q}=\frac{{b}}{{c}}\:\:\:\:\:\frac{{q}}{{p}}=\frac{{a}}{{c}} \\ $$$$\frac{{a}}{{q}}=\frac{{c}}{{p}}={r}\left({say}\right) \\ $$$${a}={rq} \\ $$$${c}={pr} \\ $$$${b}={ap}=\left({rq}\right){p}={pqr} \\ $$$${b}={qc}={q}\left({pr}\right)={pqr} \\ $$$${a}={rq}\:\:\:\:{b}={pqr}\:\:\:\:\:{c}={pr} \\ $$$$\mathrm{2}\left(\frac{{b}}{{a}}+\frac{{b}}{{c}}\right)=\frac{{a}}{{b}}+\frac{{a}}{{c}}+\frac{{c}}{{a}}+\frac{{c}}{{b}} \\ $$$$\mathrm{2}\left(\frac{{pqr}}{{rq}}+\frac{{pqr}}{{pr}}\right)=\frac{{rq}}{{pqr}}+\frac{{rq}}{{pr}}+\frac{{pr}}{{rq}}+\frac{{pr}}{{pqr}} \\ $$$$\mathrm{2}\left({p}+{q}\right)=\frac{\mathrm{1}}{{p}}+\frac{{q}}{{p}}+\frac{{p}}{{q}}+\frac{\mathrm{1}}{{q}} \\ $$$$\mathrm{2}\left({p}+{q}\right)=\frac{{p}+{q}}{{pq}}+\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }{{pq}} \\ $$$$\mathrm{2}{pq}\left({p}+{q}\right)={p}+{q}+{p}^{\mathrm{2}} +{q}^{\mathrm{2}} \\ $$$$\mathrm{2}{pq}\left({p}+{q}\right)+\mathrm{2}{pq}={p}+{q}+{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +\mathrm{2}{pq} \\ $$$$\mathrm{2}{pq}\left({p}+{q}+\mathrm{1}\right)={p}+{q}+\left({p}+{q}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{pq}\left({p}+{q}+\mathrm{1}\right)=\left({p}+{q}\right)\left(\mathrm{1}+{p}+{q}\right) \\ $$$$\left({p}+{q}+\mathrm{1}\right)\left(\mathrm{2}{pq}−{p}−{q}\right)=\mathrm{0} \\ $$$${when}\:\mathrm{2}{pq}−{p}−{q}=\mathrm{0} \\ $$$$\mathrm{2}{pq}={p}+{q}\:\left[{p}+{q}+\mathrm{1}\:{can}\:{not}\:{be}\:{equsls}\:{to}\:{zero}\right] \\ $$$$ \\ $$$${now} \\ $$$$\:\:\mathrm{2}{pa}={p}+{q} \\ $$$${multiply}\:{both}\:{side}\:{by}\:{r} \\ $$$$\mathrm{2}{pqr}={pr}+{qr} \\ $$$$\mathrm{2}{b}={c}+{a}\:\left[{since}\:{pqr}={b}\:\:\:\:\:{pr}={c}\:\:\:\:{qr}={a}\right] \\ $$$${so}\:{b}−{a}={c}−{b} \\ $$$${hence}\:{a},{b},{c}\:{are}\:{in}\:{AP} \\ $$$$ \\ $$$$ \\ $$