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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

3)2(ba+bc)=ab+ac+ca+cbp=baq=bcqp=acaq=cp=r(say)a=rqc=prb=ap=(rq)p=pqrb=qc=q(pr)=pqra=rqb=pqrc=pr2(ba+bc)=ab+ac+ca+cb2(pqrrq+pqrpr)=rqpqr+rqpr+prrq+prpqr2(p+q)=1p+qp+pq+1q2(p+q)=p+qpq+p2+q2pq2pq(p+q)=p+q+p2+q22pq(p+q)+2pq=p+q+p2+q2+2pq2pq(p+q+1)=p+q+(p+q)22pq(p+q+1)=(p+q)(1+p+q)(p+q+1)(2pq−p−q)=0when2pq−p−q=02pq=p+q[p+q+1cannotbeequslstozero]now2pa=p+qmultiplybothsidebyr2pqr=pr+qr2b=c+a[sincepqr=bpr=cqr=a]sob−a=c−bhencea,b,careinAP

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