Question

If $x$, $y$, $z$ are in H.P then the value of expression $\log (x+z)+\log (x-2y+z)$:
A) $\log \left| x-z \right|$
B) 2$\log \left| x-z \right|$
C) 3$\log \left| x-z \right|$
D) 4$\log \left| x-z \right|$

Answer 1

Hint:
We are given that $x$, $y$, $z$ are in H.P, so apply the condition of H.P. After that, take the expression $\log (x+z)+\log (x-2y+z)$ and apply the properties of log and simplify it. Try it, you will get the answer.

Complete step by step solution:
We are given that $x$, $y$, $z$ are in H.P.
Applying condition of H.P we get,
$y=\dfrac{2xz}{x+z}$ ……………. (1)
Now we have given an expression that $\log (x+z)+\log (x-2y+z)$, taking the equation and applying the property of log i.e. $\log (ab)=\log a+\log b$.
$\log (x+z)+\log (x-2y+z)=\log ((x+z)(x-2y+z))$
Now substituting the value of $y$i.e. equation (1) in the above equation.
$\log (x+z)+\log (x-2y+z)=\log ((x+z)(x+z-\dfrac{4xz}{x+z}))$
Now simplifying in simple manner we get,
$\log (x+z)+\log (x-2y+z)=\log ((x+z)(\dfrac{{{(x+z)}^{2}}-4xz}{\left( x+z \right)}))$
$\log (x+z)+\log (x-2y+z)=\log ({{(x+z)}^{2}}-4xz)$
Now we know that ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, applying the formula in above equation we get,
$\log (x+z)+\log (x-2y+z)=\log ({{x}^{2}}+{{z}^{2}}+2xz-4xz)$
$\log (x+z)+\log (x-2y+z)=\log ({{x}^{2}}+{{z}^{2}}-2xz)$
Again applying the formula we get,
$\log (x+z)+\log (x-2y+z)=\log ({{\left( x-z \right)}^{2}})$
Here, using property that $\log {{a}^{2}}=2\log a$.
$\log (x+z)+\log (x-2y+z)=2\log \left( x-z \right)$
Therefore, we get the answer as $\log (x+z)+\log (x-2y+z)=2\log \left( x-z \right)$.

Hence, the correct option is (B).

Additional information:
In Mathematics, a progression is defined as a series of numbers arranged in a predictable pattern. It is a type of number set which follows specific, definite rules. There is a difference between the progression and a sequence. A progression has a particular formula to compute its nth term, whereas a sequence is based on the specific logical rules.
A Harmonic Progression (HP) is defined as a sequence of real numbers which is determined by taking the reciprocals of the arithmetic progression that does not contain $0$. In the harmonic progression, any term in the sequence is considered as the harmonic means of its two neighbours.

Note:
Since $x$, $y$, $z$ are in H.P then $y=\dfrac{2xz}{x+z}$. Also, you must know the properties such as $\log (ab)=\log a+\log b$ and $\log {{a}^{2}}=2\log a$. We have used the formula of ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.