# 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?

Feb 18, 2017

#### Explanation:

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

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

Also $p , q , r$ are in harmonic progression and hence $\frac{1}{p} , \frac{1}{q} , \frac{1}{r}$ are in arithmetic progression, i.e. $\frac{1}{q} - \frac{1}{p} = \frac{1}{r} - \frac{1}{q}$ or

$\frac{2}{q} = \frac{1}{p} + \frac{1}{r}$ or $\frac{q}{2} = \frac{p r}{p + r}$................................(2)

As $a p , b q , c r$ are in geometric progression, hence $\frac{a p}{b q} = \frac{b q}{c r}$

or ${\left(b q\right)}^{2} = a p \times c r$ ................................(3)

Here we assume that $a p \ne c r$ i.e. $\frac{a}{c} \ne \frac{r}{p}$

Multiplying (1) and (2), we get

$b q = \left(a + c\right) \frac{p r}{p + r}$ and as $b q = \sqrt{a p c r}$, we have

$\sqrt{a p c r} = \left(a + c\right) \frac{p r}{p + r}$

or $\frac{\sqrt{a c}}{a + c} = \frac{\sqrt{p r}}{p + r}$

or $\frac{a + c}{\sqrt{a c}} = \frac{p + r}{\sqrt{p r}}$

or $\sqrt{\frac{a}{c}} + \sqrt{\frac{c}{a}} = \sqrt{\frac{p}{r}} + \sqrt{\frac{r}{p}}$

• Observe that $\sqrt{k} + \frac{1}{\sqrt{k}} = \sqrt{l} + \frac{1}{\sqrt{l}}$ is equivalent to
• $k + \frac{1}{k} = l + \frac{1}{l}$ - just square and you get it. This gives us
• $k - l = \frac{1}{l} - \frac{1}{k} = \frac{k - l}{k l}$ i.e. either $k = l$ or $k = \frac{1}{l}$

Now as $\frac{a}{c} \ne \frac{r}{p}$, we have $\frac{a}{c} = \frac{p}{r}$ or $a r = c p$

but $a r c p = {\left(b q\right)}^{2}$, hence $a r = b q = c p$

or $\frac{a}{\frac{1}{r}} = \frac{b}{\frac{1}{q}} = \frac{c}{\frac{1}{p}}$

Hence $a : b : c$ is equal to $\frac{1}{r} : \frac{1}{q} : \frac{1}{p}$