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#?

Answer 1

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#
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Answer 2

To prove that ( \frac{a}{b}:\frac{b}{q}:\frac{c}{r} = \frac{1}{r}:\frac{1}{q}:\frac{1}{p} ), we can start by expressing the terms of each progression:

Arithmetic progression: [ a = x ] [ b = x + d ] [ c = x + 2d ]

Harmonic progression: [ \frac{1}{p} = \frac{1}{x} ] [ \frac{1}{q} = \frac{1}{x + d} ] [ \frac{1}{r} = \frac{1}{x + 2d} ]

Geometric progression: [ ap = x \cdot \frac{1}{p} = 1 ] [ bq = (x + d) \cdot \frac{1}{q} = \frac{x + d}{x + d} = 1 ] [ cr = (x + 2d) \cdot \frac{1}{r} = \frac{x + 2d}{x + 2d} = 1 ]

So, ( ap = bq = cr = 1 ), which means ( ap, bq, cr ) are all equal to 1.

Therefore, ( \frac{a}{b}:\frac{b}{q}:\frac{c}{r} = 1:1:1 ), which is equivalent to ( 1/r:1/q:1/p ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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