Animated Solution for Mathematics - Sequence and Series: If $x={\sum }_{n=0}^{\infty }{a}^n,y={\sum }_{n=0}^{\infty }{b}^n,z={\sum }_{n=0}^{\infty }{c}^n$ where $a,b,c$ are in A.P and $|a|<1,|b|<1,|c|<1$ then $x,y,z$ are in

We are given the infinite series $x={\sum }_{n=0}^{\infty }{a}^n$, $y={\sum }_{n=0}^{\infty }{b}^n$, and $z={\sum }_{n=0}^{\infty }{c}^n$.

Expanding these, we see $x=1+a+a^2+a^3+\dots$, which is a classic infinite Geometric Progression (G.P.) with a first term of $1$ and a common ratio of $a$.

Given the conditions $|a|<1$, $|b|<1$, and $|c|<1$, we can apply the sum formula for an infinite G.P., which is $S_{\infty }=\frac{1}{1-r}$.

Thus, the infinite series collapse into the following compact fractions:

We are given that $a,b,$ and $c$ are in Arithmetic Progression (A.P.). This implies the relationship:

To relate this to our variables $x,y,$ and $z$, we perform a transformation. If we multiply the A.P. sequence $a,b,c$ by $-1$, we obtain $-a,-b,-c$, which remains in A.P.

Adding $1$ to each term of this sequence yields $1-a,1-b,1-c$. Since adding a constant to every term of an A.P. preserves the A.P. property, the sequence $1-a,1-b,1-c$ is also in A.P.

From our earlier expressions, we know that $1-a=\frac{1}{x}$, $1-b=\frac{1}{y}$, and $1-c=\frac{1}{z}$.

Since $1-a,1-b,1-c$ are in A.P., it follows that their reciprocals must satisfy the condition for an Arithmetic Progression:

By definition, a sequence is in Harmonic Progression (H.P.) if the reciprocals of its terms form an A.P.

Therefore, we conclude that the sequence $x,y,z$ is in Harmonic Progression (H.P.).