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# If log(p+r)+log(p+r-2q)=2log(p-r) then p, q and r are in A. APB. GPC. HPD. None of these

Last updated date: 09th Aug 2024
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Hint: In order to solve this problem you need to use the property of logarithm. That is log(ab)=loga + logb and $a\log b = \log {b^a}$ then we will be getting the equations solving those equations and knowing the properties of HP you will get the right answer.

It is given that, log(p+r)+log(p+r-2q)=2log(p-r)
We have to use property of logarithm log(ab)=loga + logb and $a\log b = \log {b^a}$
Now, using above identity we can write log(p+r)+log(p+r-2q)=2log(p-r) as
$\log ((p + r)(p + r - 2q)) = \log {(p - r)^2}$
On cancelling log from both sides we get the equation as,
$\Rightarrow (p + r)(p + r - 2q) = {(p - r)^2} \\ \Rightarrow {p^2} + pr - 2qp + pr + {r^2} - 2pr \\$
Now, ${p^2}$ and ${r^2}$ are cancelled,
Then the equation will become,
2pr – 2pq – 2qr = -2pr
4pr = 2pq + 2qr
On dividing this equation by 2pqr we get,
$\Rightarrow \dfrac{2}{q} = \dfrac{1}{p} + \dfrac{1}{r}$
This condition shows that p, q, r makes an HP where the above condition is the harmonic mean of the HP p, q, r.
Now, we can easily see p, q, r is in HP.

So, the correct answer is “Option C”.

Note: Whenever we face such types of problems we use some important points. Like always trying to remove log from both sides of the equation by using the identity of logarithm then after some easy rearrangement we can easily identify which type of progression formed.