Problem on infinite geometric sequence

The sum of an infinite geometric sequence is $\frac{a_1}{1-r}$ , and it is known that this sum is negative, so $\frac{a_1}{1-r} < 0$ , so either the numerator or the denominator must be negative (and the other one must be positive, of course). If so, the product of the numerator and the denominator is also necessarily negative: $a_1\left(1-r\right)<0$ . Since the sum of the sequence is convergent, that means $-1<r<1\Leftrightarrow 0<1-r<2$ . Since $1-r$ is always positive, we can divide both sides of the inequality by it and be left with $a_1<0$ . This is the correct answer.