A) \[\log (x-z)\]
B) \[2\log (x-z)\]
C) \[3\log (x-z)\]
D) \[4\log (x-z)\]
Correct Answer: B
Solution :
If \[x,\ y,\,z\] are in H.P., then \[y=\frac{2xz}{x+z}\] Now, \[{{\log }_{e}}(x+z)+{{\log }_{e}}(x-2y+z)\] \[\sum\limits_{i=1}^{n}{{}}=3\sum\limits_{i=1}^{n}{i}-2\sum\limits_{i=1}^{n}{1}=3\frac{n(n+1)}{2}-2n=\frac{n(3n-1)}{2}\] \[={{\log }_{e}}\left[ (x+z)\,\left( x+z-\frac{4xz}{x+z} \right) \right]\] \[={{\log }_{e}}[{{(x+z)}^{2}}-4xz]={{\log }_{e}}{{(x-z)}^{2}}=2{{\log }_{e}}(x-z)\].
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