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(B) The sum of an infinite geometric progression is given by $S = \frac{a}{1-r}$,where $a$ is the first term and $r$ is the common ratio,with the cond

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$0 < x < 10$

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(B) The sum of an infinite geometric progression is given by $S = \frac{a}{1-r}$,where $a$ is the first term and $r$ is the common ratio,with the condition $|r| Given $a = x$ and $S = 5$,we have $5 = \frac{x}{1-r}$.Rearranging for $r$,we get $1 - r = \frac{x}{5}$,which implies $r = 1 - \frac{x}{5}$.Since $|r| This inequality can be written as $-1 Subtracting $1$ from all parts,we get $-2 Multiplying by $-5$ (and reversing the inequality signs),we get $10 > x > 0$,or $0 < x < 10$.

The first term of an infinite geometric progression is $x$ and its sum is $5$. Then:

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