1. Evaluate the given sums

Since $|a|<1$, $|b|<1$, $|c|<1$, each series is a convergent geometric series:

$x=\sum_{n=0}^{\infty} a^n=\frac{1}{1-a},\qquad y=\sum_{n=0}^{\infty} b^n=\frac{1}{1-b},\qquad z=\sum_{n=0}^{\infty} c^n=\frac{1}{1-c}.$

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

If $a,b,c$ are in arithmetic progression, then

$2b=a+c.$

We need to determine the relation among $x,y,z$.

3. Take reciprocals

From the expressions above,

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

Now check whether these are in A.P.:

$\frac{1}{x}+\frac{1}{z}=(1-a)+(1-c)=2-(a+c).$

Since $a+c=2b$, this becomes

$\frac{1}{x}+\frac{1}{z}=2-2b=2(1-b)=\frac{2}{y}.$

So,

$2\cdot \frac{1}{y}=\frac{1}{x}+\frac{1}{z}.$

Hence $\dfrac1x,\dfrac1y,\dfrac1z$ are in A.P.

4. Conclusion

If the reciprocals of three numbers are in A.P., then the numbers themselves are in H.P. Therefore,

$x,y,z \text{ are in H.P.}$

5. Check options

• A: G.P. $\rightarrow$ Not necessarily
• B: A.P. $\rightarrow$ Not necessarily
• C: Arithmetic-Geometric Progression $\rightarrow$ No
• D: H.P. $\rightarrow$ Correct

Therefore, the correct answer is D.