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 $n.$
Note that if the the two series have constant ratios of $a$ and $b,$ respectively, then $4\left( \frac{12}{1-a} \right) = \frac{12}{1-b}.$ Simplifying, $4(1-b)=1-a.$ Substituting in $a= \frac{4}{12}=\frac{1}{3}$ and $b= \frac{4+n}{12}=\frac{1}{3}+\frac{n}{12},$ we quickly find that $n=\boxed{6}.$
6
Level 5
Algebra
Problem: 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 $n.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for \( n \), we first need to determine the common ratio of the first infinite geometric series. The first term is \( a_1 = 12 \) and the second term is \( a_2 = 4 \). The common ratio \( r_1 \) can be found using the formula for the second term of a geometric series, which is \( a_2 = a_1 \cdot r_1 \). Thus, we have:

\[
4 = 12 \cdot r_1
\]

Solving for \( r_1 \):

\[
r_1 = \frac{4}{12} = \frac{1}{3}
\]

Next, we calculate the sum \( S_1 \) of the first infinite geometric series using the formula \( S = \frac{a_1}{1 - r} \):

\[
S_1 = \frac{12}{1 - \frac{1}{3}} = \frac{12}{\frac{2}{3}} = 12 \cdot \frac{3}{2} = 18
\]

Now, we turn to the second infinite geometric series, which has the same first term \( a_1 = 12 \) and a second term \( a_2 = 4 + n \). Let the common ratio of the second series be \( r_2 \). We can express \( r_2 \) as follows:

\[
4 + n = 12 \cdot r_2
\]

From this, we can solve for \( r_2 \):

\[
r_2 = \frac{4 + n}{12}
\]

The sum \( S_2 \) of the second series is given to be four times the sum of the first series:

\[
S_2 = 4 \cdot S_1 = 4 \cdot 18 = 72
\]

Using the sum formula for the second series:

\[
S_2 = \frac{12}{1 - r_2}
\]

Setting this equal to 72 gives us:

\[
\frac{12}{1 - r_2} = 72
\]

Cross-multiplying yields:

\[
12 = 72(1 - r_2)
\]

Expanding and rearranging:

\[
12 = 72 - 72r_2
\]
\[
72r_2 = 72 - 12
\]
\[
72r_2 = 60
\]
\[
r_2 = \frac{60}{72} = \frac{5}{6}
\]

Now we substitute \( r_2 \) back into the equation for \( r_2 \):

\[
\frac{4 + n}{12} = \frac{5}{6}
\]

Cross-multiplying gives:

\[
4 + n = 12 \cdot \frac{5}{6}
\]
\[
4 + n = 10
\]

Solving for \( n \):

\[
n = 10 - 4 = 6
\]

Thus, the value of \( n \) is \( 6 \).<|END_OF_LATENT|> The final answer is $6$.