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 $1/r:1/q:1/p$ ?

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Answer 1 · 2017-02-18T09:23:47

Please see below.

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 $sqrt(a/c)+sqrt(c/a)=sqrt(p/r)+sqrt(r/p)$

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 $1/r:1/q:1/p$

If a,b,ca,b,c are in arithmetic progression; p,q,rp,q,r are in harmonic progression; ap,bq,crap,bq,cr are in geometric progression then prove that a:b:ca:b:c is equal to 1/r:1/q:1/p1/r:1/q:1/p?