If #a,b,c# are in arithmetic progression; #p,q,r# are in harmonic progression; #ap,bq,cr# are in geometric progression then prove that #a:b:c# is equal to #1/r:1/q:1/p#?

1 Answer
Feb 18, 2017

Answer:

Please see below.

Explanation:

As #a,b,c# are in arithmetic progression, we have #b-a=c-b#

or #2b=a+c# ................................(1)

Also #p,q,r# are in harmonic progression and hence #1/p,1/q,1/r# are in arithmetic progression, i.e. #1/q-1/p=1/r-1/q# or

#2/q=1/p+1/r# or #q/2=(pr)/(p+r)#................................(2)

As #ap,bq,cr# are in geometric progression, hence #(ap)/(bq)=(bq)/(cr)#

or #(bq)^2=apxxcr# ................................(3)

Here we assume that #ap!=cr# i.e. #a/c!=r/p#

Multiplying (1) and (2), we get

#bq=(a+c)(pr)/(p+r)# and as #bq=sqrt(apcr)#, we have

#sqrt(apcr)=(a+c)(pr)/(p+r)#

or #sqrt(ac)/(a+c)=sqrt(pr)/(p+r)#

or #(a+c)/sqrt(ac)=(p+r)/sqrt(pr)#

or #sqrt(a/c)+sqrt(c/a)=sqrt(p/r)+sqrt(r/p)#

  • Observe that #sqrtk+1/sqrtk=sqrtl+1/sqrtl# is equivalent to
  • #k+1/k=l+1/l# - just square and you get it. This gives us
  • #k-l=1/l-1/k=(k-l)/(kl)# i.e. either #k=l# or #k=1/l#

Now as #a/c!=r/p#, we have #a/c=p/r# or #ar=cp#

but #arcp=(bq)^2#, hence #ar=bq=cp#

or #a/(1/r)=b/(1/q)=c/(1/p)#

Hence #a:b:c# is equal to #1/r:1/q:1/p#