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 $n.$
your first series has
a = 12, r = 4/12 = 1/3
and the sum of all the terms is 12/(1-1/3) = 18
your second has a = 12, and r = (4+n)/12
the sum of all the terms = 12/(1 - (4+n)/12)
= 12/( (12 - 4 - n)/12)
= 144/(8 - n)
but you told us that this is 4(18)
144/(8-n) = 72
144 = 576 - 72n
72n = 432
n = 6
To find the value of $n,$ we can start by finding the common ratio of the first infinite geometric series.
Let's call the common ratio of the first series $r.$ The first term is $12$ and the second term is $4,$ so we can set up the following equation:
$$12\cdot r = 4.$$
Simplifying, we have:
$$r = \frac{4}{12} = \frac{1}{3}.$$
The first series has a common ratio of $\frac{1}{3}.$
Now, let's find the sum of the first series. The formula to find the sum of an infinite geometric series is:
$$\text{Sum} = \frac{\text{First Term}}{1 - \text{Common Ratio}}.$$
Substituting the values we have, we get:
$$\text{Sum}_1 = \frac{12}{1-\frac{1}{3}} = \frac{12}{\frac{2}{3}} = \frac{12\cdot 3}{2} = 18.$$
Now, let's find the second term of the second series. We know that it is $4+n.$
Since the first term of the second series is the same as the first term of the first series, we can use the same common ratio, $\frac{1}{3}.$
Therefore, the second term of the second series is:
$$12 \cdot \frac{1}{3} = 4.$$
We are given that the sum of the second series is four times the sum of the first series. Let's call the sum of the second series $\text{Sum}_2.$
So we have:
$$\text{Sum}_2 = 4 \cdot \text{Sum}_1 = 4 \cdot 18 = 72.$$
Finally, we can find the common ratio of the second series:
$$12 \cdot r_2 = 4 + n.$$
Simplifying, we have:
$$r_2 = \frac{4+n}{12}.$$
The sum of the second series is four times the sum of the first series, so we can set up the equation:
$$\text{Sum}_2 = \frac{12}{1 - r_2} = 72.$$
Substituting the value of $\text{Sum}_2$ and $r_2,$ we have:
$$\frac{12}{1 - \frac{4+n}{12}} = 72.$$
Multiplying both sides by $1 - \frac{4+n}{12},$ we get:
$$12 = 72\left(1 - \frac{4+n}{12}\right).$$
Simplifying, we have:
$$1 - \frac{4+n}{12} = \frac{12}{72} = \frac{1}{6}.$$
Multiplying both sides by $12,$ we have:
$$12 - 4 - n = 2.$$
Simplifying, we find:
$$n = 12 - 4 - 2 = \boxed{6}.$$