+230 An infinite geometric series has a first term of $12$ and a second term of $4$. A second infinite geometric series has the same first term of $12$ a second term of $4+n,$ and a sum of four times that of the first series. Find the value of $n$.
+1223 THe first geometric series has first term 12 and common ratio 1/3, so we can find the sum.
\(S = \frac{a}{1-r} = \frac{12}{1-\frac{1}{3}} = 18\) . (note that this formula only works if the absolute value of the ratio is less than 1)
For the second series, we know that the first term is 12 and the common ratio is (4+n)/12. The sum is also 4(18) = 72, so we write:
\(72 = \frac{12}{1 - \frac{4+n}{12}}\)
Solving for n, we get \(\boxed{n=6}\)
