2016 AMC 12B Problems/Problem 14
Contents
Problem
The sum of an infinite geometric series is a positive number , and the second term in the series is
. What is the smallest possible value of
Solution
The second term in a geometric series is , where
is the common ratio for the series and
is the first term of the series. So we know that
and we wish to find the minimum value of the infinite sum of the series. We know that:
and substituting in
, we get that
. From here, you can either use calculus or AM-GM.
Let , then
. Since
and
are undefined
. This means that we only need to find where the derivative equals
, meaning
. So
, meaning that
For 2 positive real numbers and
,
. Let
and
. Then:
. This implies that
. or
. Rearranging :
. Thus, the smallest value is
.
Solution 1
The sum of the geometric sequence is where
is the first term and
is the common ratio. We know the second term,
is equal to
Thus
This means,
$$ (Error compiling LaTeX. Unknown error_msg)S = \frac{a}{1 - r} = \frac{1/r}{1 - r} = \frac{r(1 - r)}.
S,
r = 1-r \Rightarrow r = \frac{1}{2}.
S
\frac{\frac{1}{1/2}}{1 - \frac{1}{2}} = \boxed{(E)~4}.$==Solution 2==
A geometric sequence always looks like
<cmath>a,ar,ar^2,ar^3,\dots</cmath>
and they say that the second term$ (Error compiling LaTeX. Unknown error_msg)ar=1S
\frac{a}{1-r}
S$in one variable.
<cmath> \begin{align*} ar&=1 \\ S&=\frac{a}{1-r} \end{align*} </cmath>$ (Error compiling LaTeX. Unknown error_msg)\textbf{Solving in terms of \textit{a} then graphing}$<cmath>S=\frac{a^2}{a-1}</cmath>
We seek the smallest positive value of$ (Error compiling LaTeX. Unknown error_msg)SaS
\boxed{\textbf{(E)}\ 4}.$$ (Error compiling LaTeX. Unknown error_msg)\textbf{Solving in terms of \textit{r} then graphing}$<cmath>S=\frac{1}{-r^2+r}</cmath>
We seek the smallest positive value of$ (Error compiling LaTeX. Unknown error_msg)SrS
\boxed{\textbf{(E)}\ 4}.$$ (Error compiling LaTeX. Unknown error_msg)\textbf{Solving in terms of \textit{a} then doing some calculus}$<cmath>S=\frac{a^2}{a-1}</cmath>
We seek the smallest positive value of$ (Error compiling LaTeX. Unknown error_msg)S\frac{(a-2)a}{(a-1)^2}=S'
\frac{(a-2)a}{(a-1)^2}=0
a=0
a=2
\frac{2}{(a-1)^3}=S
\frac{2}{(0-1)^3}
a=0
\frac{2}{(2-1)^3}
a=2
a=2
S=4
S
\boxed{\textbf{(E)}\ 4}.
S
S
S
\textbf{Solving in terms of \textit{r} then being clever}$<cmath>S=\frac{1}{-r^2+r}</cmath>
We seek the smallest positive value of$ (Error compiling LaTeX. Unknown error_msg)Sx=\frac{1}{2}
\frac{1}{-(\frac{1}{2})^2+\frac{1}{2}}=\boxed{\textbf{(E)}\ 4}$.
==Solution 3==
<cmath>\textbf{Completing the Square and Quadratics}</cmath>
Let$ (Error compiling LaTeX. Unknown error_msg)r1
\frac{1}{r}
\frac{1}{r(1-r)}
r(1-r)
\frac{1}{4}
S
\boxed{\textbf{(E)}\ 4}
a_1,1,...
(0,1)
\frac{1}{a_1}
S
\frac{a_1}{1-\frac{1}{a_1}}
S=\frac{1}{a_1-1}
a_1
2$(makes S positive & we know a sequence/series of a ratio$ (Error compiling LaTeX. Unknown error_msg)1/2
2,1,1/2,1,4,...
2+1+1=\boxed{\textbf{(E)}\ 4}$.
-thedodecagon
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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