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Q.
If $x$ is a positive real number different from $1,$ then the number $\frac{1}{1+\sqrt{x}}, \frac{1}{1-x}, \frac{1}{1-\sqrt{x}}, \ldots$ are in
Sequences and Series
Solution:
Since $\frac{1}{1-x}-\frac{1}{1+\sqrt{x}}=\frac{1-(1-\sqrt{x})}{1-x}=\frac{\sqrt{x}}{1-x}$
And $\frac{1}{1-\sqrt{x}}-\frac{1}{1-x}=\frac{1+\sqrt{x}-1}{1-x}=\frac{\sqrt{x}}{1-x}$
Therefore, given numbers are in $A.P$.