An infinite geometric series has common ratio \(-1/5\) and sum \(16.\) What is the first term of the series?
Sum of a geometric series =
a1 / [ 1 - r ] where a1 is the first term and r is the common ratio....so we have
16 = a1 / [ 1 - -1/5]
16 = a1 / [ 6/5] multiply both sides by 6/5
(6/5)16 = a1 =
96/5
Solution: Let the first term be \(a\). Because the sum of the series is \(16\) , we have \(16= \frac{a}{1-(-1/5)} = \frac{a}{6/5} = \frac{5a}{6}\) . Therefore, \(a=\boxed{\frac{96}{5}}\) .
