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Leta, band c be in AP and $|a| < 1,|b| < 1,|c| < 1$.
If $x=1+a+a^{2}+\ldots$ to $\infty$
$\mathrm{y}=1+\mathrm{b}+\mathrm{b}^{2}+\ldots$ to $\infty$
$\mathrm{z}=1+\mathrm{c}+\mathrm{c}^{2}+\ldots$ to $\infty$, then $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are in
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If $x=1+a+a^{2}+\ldots$ to $\infty$
$\mathrm{y}=1+\mathrm{b}+\mathrm{b}^{2}+\ldots$ to $\infty$
$\mathrm{z}=1+\mathrm{c}+\mathrm{c}^{2}+\ldots$ to $\infty$, then $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are in
Solution:
1331 Upvotes
Verified Answer
The correct answer is:
$\mathrm{HP}$
$$
\begin{array}{l}
x=1+a+a^{2}+\ldots \infty=\frac{1}{1-a} \\
y=1+b+b^{2}+\ldots \infty=\frac{1}{1-b}
\end{array}
$$
and $z=1+c+c^{2}+\ldots \infty=\frac{1}{1-c}$
Since, a, b and c are in AP.
$\Rightarrow 1-a, 1-b$ and $1-c$ are also in $A P$
$\Rightarrow \frac{1}{1-a}, \frac{1}{1-b}$ and $\frac{1}{1-c}$ are in HP.
$\therefore \mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are in $\mathrm{HP}$.
Note that if the common ratio of a GP is not less than 1 , then we do not determ ined the sum of an infinite GP that series.
\begin{array}{l}
x=1+a+a^{2}+\ldots \infty=\frac{1}{1-a} \\
y=1+b+b^{2}+\ldots \infty=\frac{1}{1-b}
\end{array}
$$
and $z=1+c+c^{2}+\ldots \infty=\frac{1}{1-c}$
Since, a, b and c are in AP.
$\Rightarrow 1-a, 1-b$ and $1-c$ are also in $A P$
$\Rightarrow \frac{1}{1-a}, \frac{1}{1-b}$ and $\frac{1}{1-c}$ are in HP.
$\therefore \mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are in $\mathrm{HP}$.
Note that if the common ratio of a GP is not less than 1 , then we do not determ ined the sum of an infinite GP that series.
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