In a geometric progression, the first term is $a$ (with $a \ne 0$), the common ratio is $r$ (with $r < 1$), and the sum to infinity is $S$. If the second term of this geometric progression is equal to the negative of the sum to infinity, what is the value of $[r]$, where $[r]$ denotes the greatest integer function?

For a geometric progression with first term $a$ and common ratio $r$ where $|r| < 1$, the sum to infinity is:

$S = \frac{a}{1-r}$

The second term of the geometric progression is $ar$.

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Given that the second term equals the negative of the sum to infinity:

$ar = -S$

Substituting the formula for $S$:

$ar = -\frac{a}{1-r}$

Since $a \ne 0$, dividing both sides by $a$:

$r = -\frac{1}{1-r}$

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$r(1-r) = -1$

$r - r^2 = -1$

$r^2 - r - 1 = 0$

Using the quadratic formula:

$r = \frac{1 \pm \sqrt{1 + 4}}{2}$

$r = \frac{1 \pm \sqrt{5}}{2}$

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The two solutions are:

$r = \frac{1 + \sqrt{5}}{2} \approx 1.618$

$r = \frac{1 - \sqrt{5}}{2} \approx -0.618$

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Note: The question says $r < 1$, but not $|r| < 1$ (that is, $r$ between $-1$ and $1$). This matters because the sum to infinity exists only when $|r| < 1$. Still, since the question says the sum to infinity $S$ exists, we can take $|r| < 1$ as given.

Either way, $r \approx 1.618$ is rejected and $r \approx -0.618$ is accepted. So the valid solution is:

$r = \frac{1 - \sqrt{5}}{2} \approx -0.618$

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Since $\sqrt{5} \approx 2.236$:

$r = \frac{1 - 2.236}{2} = \frac{-1.236}{2} \approx -0.618$

This means $-1 < r < 0$.

The greatest integer function $[r]$ gives the largest integer less than or equal to $r$.

Therefore, $[r] = -1$

The answer is Option 4: $-1$