Find the sum of the series \(96 - 48 + 24 - 12 + ...\) up to infinity.
Answer: A
This is an infinite G.P. with first term \(a=96\) and common ratio \(r = -48/96 = -1/2\). Since \(|r|<1\), the sum to infinity exists. \(S_\infty = \frac{a}{1-r} = \frac{96}{1 - (-1/2)} = \frac{96}{1+1/2} = \frac{96}{3/2} = 96 \times \frac{2}{3} = 32 \times 2 = 64\).