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**Hint:**We start solving the problem by expanding the given summations of x, y and z. After expanding the summation, we can see that the sum resembles the sum of infinite geometric series. So, we use the sum of the infinite geometric series $\dfrac{p}{1-r}$ if $\left| r \right|<1$ to get the a, b and c in terms of x, y and z. We then use the information given about a, b and c in order to get the desired relation between x, y and z.

__Complete step-by-step answer__:According to the problem, we are given $x=\sum\limits_{n=0}^{\infty }{{{a}^{n}}}$, $y=\sum\limits_{n=0}^{\infty }{{{b}^{n}}}$, $z=\sum\limits_{n=0}^{\infty }{{{c}^{n}}}$ where a, b, c are in A.P and $\left| a \right|<1$, $\left| b \right|<1$, $\left| c \right|<1$. We need to find the progression that the numbers x, y, z are in.

Let us first solve for the values of x, y and z.

We have $x=\sum\limits_{n=0}^{\infty }{{{a}^{n}}}$.

$\Rightarrow x={{a}^{0}}+{{a}^{1}}+{{a}^{2}}+{{a}^{3}}+..............\infty $.

$\Rightarrow x=1+a+{{a}^{2}}+{{a}^{3}}+..............\infty $ ---(1).

We can see that the right-hand side resembles the sum of infinite geometric series with the first term as 1 and the common ratio as ‘a’. According the problem we have $\left| a \right|<1$.

We know that the sum of the terms in infinite geometric series $p+pr+p{{r}^{2}}+p{{r}^{3}}+.......\infty $ is $\dfrac{p}{1-r}$ if $\left| r \right|<1$. We use this result in equation (1).

$\Rightarrow x=\dfrac{1}{1-a}$.

$\Rightarrow 1-a=\dfrac{1}{x}$ ---(2).

Now, we have $y=\sum\limits_{n=0}^{\infty }{{{b}^{n}}}$.

$\Rightarrow y={{b}^{0}}+{{b}^{1}}+{{b}^{2}}+{{b}^{3}}+..............\infty $.

$\Rightarrow y=1+b+{{b}^{2}}+{{b}^{3}}+..............\infty $ ---(3).

We can see that the right-hand side resembles the sum of infinite geometric series with first term as 1 and the common ratio as ‘b’. According the problem we have $\left| b \right|<1$.

We know that the sum of the terms in infinite geometric series $p+pr+p{{r}^{2}}+p{{r}^{3}}+.......\infty $ is $\dfrac{p}{1-r}$ if $\left| r \right|<1$. We use this result in equation (3).

$\Rightarrow y=\dfrac{1}{1-b}$.

$\Rightarrow 1-b=\dfrac{1}{y}$ ---(4).

Now, we have $z=\sum\limits_{n=0}^{\infty }{{{c}^{n}}}$.

$\Rightarrow z={{c}^{0}}+{{c}^{1}}+{{c}^{2}}+{{c}^{3}}+..............\infty $.

$\Rightarrow z=1+c+{{c}^{2}}+{{c}^{3}}+..............\infty $ ---(5).

We can see that the right-hand side resembles the sum of infinite geometric series with the first term as 1 and the common ratio as ‘c’. According the problem we have$\left| c \right|<1$.

We know that the sum of the terms in infinite geometric series $p+pr+p{{r}^{2}}+p{{r}^{3}}+.......\infty $ is $\dfrac{p}{1-r}$ if $\left| r \right|<1$. We use this result in equation (5).

$\Rightarrow z=\dfrac{1}{1-c}$.

$\Rightarrow 1-c=\dfrac{1}{z}$ ---(6).

According to the problem, we have a, b and c are in A.P (Arithmetic Progression). We know that if the given terms are in A.P (Arithmetic Progression) and all the terms are multiplied or divided or added or subtracted with any number, then the new terms we obtained will also be in A.P(Arithmetic Progression). Using this fact, we get the numbers –a, –b and –c are in A.P. Also, the numbers $1-a,1-b,1-c$ are also in A.P.

So, we have got 1 – a, 1 – b, 1 – c are in A.P which makes $\dfrac{1}{x}$, $\dfrac{1}{y}$, $\dfrac{1}{z}$ are in A.P.

We know that the inverse of $\dfrac{1}{x}$, $\dfrac{1}{y}$, $\dfrac{1}{z}$ are x, y, z.

We know that if p, q and r are in A.P, then $\dfrac{1}{p}$, $\dfrac{1}{q}$ and $\dfrac{1}{r}$ are in H.P (Harmonic Progression). Using this fact, we get that x, y, z are in H.P.

∴ We have found x, y, z are in H.P.

**The correct option for the given problem is (a).**

**Note**: We can prove that when the given terms are in A.P (Arithmetic Progression) and all the terms are multiplied or divided or added or subtracted with any number, then the new terms we obtained will also be in A.P(Arithmetic Progression) by taking example of any series that resembles A.P(Arithmetic Progression). We should not confuse the inverse of A.P as G.P. Similarly, we can expect problems in which x, y and z are in G.P.

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