1. Evaluate the infinite sums

Given $x=\sum_{n=0}^{\infty} a^n,\qquad y=\sum_{n=0}^{\infty} b^n,\qquad z=\sum_{n=0}^{\infty} c^n$ with $|a|,|b|,|c|<1$, each is a convergent geometric series.

So, $x=\frac{1}{1-a},\qquad y=\frac{1}{1-b},\qquad z=\frac{1}{1-c}.$

Hence, $\frac1x=1-a,\qquad \frac1y=1-b,\qquad \frac1z=1-c.$

2. Use the condition that $a,b,c$ are in A.P.

Since $a,b,c$ are in arithmetic progression, $2b=a+c.$

Now consider $\frac1x=1-a,\quad \frac1y=1-b,\quad \frac1z=1-c.$

Check whether these are in A.P.: $2\left(\frac1y\right)=2(1-b)=2-2b,$ and $\frac1x+\frac1z=(1-a)+(1-c)=2-(a+c).$

But $a+c=2b$, so $\frac1x+\frac1z=2-2b=2\left(\frac1y\right).$

Therefore, $\frac1x,\frac1y,\frac1z$ are in A.P.

So Option C is correct.

3. Check the other options

Option A: $x,y,z$ are in A.P.

This would require $2y=x+z,$ i.e. $\frac{2}{1-b}=\frac{1}{1-a}+\frac{1}{1-c}.$ This is not true in general.

Counterexample: take $a=0.1, b=0.2, c=0.3$ (which are in A.P.). Then $x=\frac1{0.9},\quad y=\frac1{0.8},\quad z=\frac1{0.7}.$ Numerically, $x\approx1.111,\ y=1.25,\ z\approx1.429.$ Then $2y=2.5,\qquad x+z\approx2.54,$ not equal.

So A is false.

Option B: $x,y,z$ are in G.P.

This would require $y^2=xz,$ i.e. $\frac{1}{(1-b)^2}=\frac{1}{(1-a)(1-c)}.$ Equivalently, $(1-b)^2=(1-a)(1-c),$ which is not generally true.

Using the same example $a=0.1,b=0.2,c=0.3$, $y^2=(1.25)^2=1.5625,$ while $xz\approx1.111\times1.429\approx1.587.$ Not equal.

So B is false.

Option D: $\frac1x+\frac1y+\frac1z=1-(a+b+c)$

But $\frac1x+\frac1y+\frac1z=(1-a)+(1-b)+(1-c)=3-(a+b+c).$ This is not equal to $1-(a+b+c)$.

So D is false.

4. Conclusion

The only correct option is $\boxed{\text{C}}.$