Found problems: 150
$(a_i),(b_i)$ are nonconstant arithmetic and geometric progressions.
$a_1=b_1,a_2/b_2=2,a_4/b_4=8$ Find $a_3/b_3$.
Let $a_1, \ldots, a_n$ be $n$ positive numbers and $0 < q < 1.$ Determine $n$ positive numbers $b_1, \ldots, b_n$ so that:
[i]a.)[/i] $ a_{k} < b_{k}$ for all $k = 1, \ldots, n,$
[i]b.)[/i] $q < \frac{b_{k+1}}{b_{k}} < \frac{1}{q}$ for all $k = 1, \ldots, n-1,$
[i]c.)[/i] $\sum \limits^n_{k=1} b_k < \frac{1+q}{1-q} \cdot \sum \limits^n_{k=1} a_k.$
Find the maximal constant $ M$, such that for arbitrary integer $ n\geq 3,$ there exist two sequences of positive real number $ a_{1},a_{2},\cdots,a_{n},$ and $ b_{1},b_{2},\cdots,b_{n},$ satisfying
(1):$ \sum_{k \equal{} 1}^{n}b_{k} \equal{} 1,2b_{k}\geq b_{k \minus{} 1} \plus{} b_{k \plus{} 1},k \equal{} 2,3,\cdots,n \minus{} 1;$
(2):$ a_{k}^2\leq 1 \plus{} \sum_{i \equal{} 1}^{k}a_{i}b_{i},k \equal{} 1,2,3,\cdots,n, a_{n}\equiv M$.
The first three terms of a geometric sequence are the integers $a,\,720,$ and $b,$ where $a<720<b.$ What is the sum of the digits of the least possible value of $b?$
$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 21$
Consider $ 2011 $ positive real numbers $ a_1,a_2,\ldots ,a_{2011} . $ If they are in geometric progression, show that there exists a real number $ \lambda $ such that any $ i\in\{ 1,2,\ldots , 1005 \} $ implies $ \lambda =a_i\cdot a_{2012-i} . $ Disprove the converse.
[i]Teodor Radu[/i]
Find all three digit natural numbers of the form $(abc)_{10}$ such that $(abc)_{10}$, $(bca)_{10}$ and $(cab)_{10}$ are in geometric progression. (Here $(abc)_{10}$ is representation in base $10$.)
Find all triples of natural numbers $(a,b,c)$ such that $a$, $b$ and $c$ are in geometric progression and $a + b + c = 111$.
Positive integers $ a$, $ b$, and $ 2009$, with $ a<b<2009$, form a geometric sequence with an integer ratio. What is $ a$?
$ \textbf{(A)}\ 7 \qquad
\textbf{(B)}\ 41 \qquad
\textbf{(C)}\ 49 \qquad
\textbf{(D)}\ 289 \qquad
\textbf{(E)}\ 2009$
A sequence of three real numbers forms an arithmetic progression with a first term of $ 9$. If $ 2$ is added to the second term and $ 20$ is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term in the geometric progression?
$ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 49 \qquad \textbf{(E)}\ 81$
Let the numbers $a_1,a_2,a_3,a_4$ form an arithmetic progression with difference $d\ne0$. Prove that there are no exists geometric progressions $b_1,b_2,b_3,b_4$ and $c_1,c_2,c_3,c_4$ such that:
$$a_1=b_1+c_1,a_2=b_2+c_2,a_3=b_3+c_3,a_4=b_4+c_4.$$
Can the numbers $1,2,3,\ldots,100$ be covered with $12$ geometric progressions?
[i]A. Golovanov[/i]
Decide whether it is possible to color the $1984$ natural numbers $1, 2, 3, \cdots, 1984$ using $15$ colors so that no geometric sequence of length $3$ of the same color exists.
Let $a_1,a_2,...$ be an arithmetic sequence with the common difference between terms is positive. Assume there are $k$ terms of this sequence creates an geometric sequence with common ratio $d$. Prove that $n\ge 2^{k-1}$.
In the adjoining figure the five circles are tangent to one another consecutively and to the lines $L_1$ and $L_2$ ($L_1$ is the line that is above the circles and $L_2$ is the line that goes under the circles). If the radius of the largest circle is 18 and that of the smallest one is 8, then the radius of the middle circle is
[asy]
size(250);defaultpen(linewidth(0.7));
real alpha=5.797939254, x=71.191836;
int i;
for(i=0; i<5; i=i+1) {
real r=8*(sqrt(6)/2)^i;
draw(Circle((x+r)*dir(alpha), r));
x=x+2r;
}
real x=71.191836+40+20*sqrt(6), r=18;
pair A=tangent(origin, (x+r)*dir(alpha), r, 1), B=tangent(origin, (x+r)*dir(alpha), r, 2);
pair A1=300*dir(origin--A), B1=300*dir(origin--B);
draw(B1--origin--A1);
pair X=(69,-5), X1=reflect(origin, (x+r)*dir(alpha))*X,
Y=(200,-5), Y1=reflect(origin, (x+r)*dir(alpha))*Y,
Z=(130,0), Z1=reflect(origin, (x+r)*dir(alpha))*Z;
clip(X--Y--Y1--X1--cycle);
label("$L_2$", Z, S);
label("$L_1$", Z1, dir(2*alpha)*dir(90));[/asy]
$\text{(A)} \ 12 \qquad \text{(B)} \ 12.5 \qquad \text{(C)} \ 13 \qquad \text{(D)} \ 13.5 \qquad \text{(E)} \ 14$
A positive integer is a good number, if its base $10$ representation can be split into at least $5$ sections, each section with a non-zero digit, and after interpreting each section as a positive integer (omitting leading zero digits), they can be split into two groups, such that each group can be reordered to form a geometric sequence (if a group has $1$ or $2$ numbers, it is also a geometric sequence), for example $20240327$ is a good number, since after splitting it as $2|02|403|2|7$, $2|02|2$ and $403|7$ form two groups of geometric sequences.
If $a>1$, $m>2$, $p=1+a+a^2+\dots+a^m$ is a prime, prove that $\frac{10^{p-1}-1}{p}$ is a good number.
Consider a function $f$ on nonnegative integers such that $f(0)=1, f(1)=0$ and $f(n)+f(n-1)=nf(n-1)+(n-1)f(n-2)$ for $n \ge 2$. Show that
\[\frac{f(n)}{n!}=\sum_{k=0}^n \frac{(-1)^k}{k!}\]
In a rectangular plot of land, a man walks in a very peculiar fashion. Labeling the corners $ABCD$, he starts at $A$ and walks to $C$. Then, he walks to the midpoint of side $AD$, say $A_1$. Then, he walks to the midpoint of side $CD$ say $C_1$, and then the midpoint of $A_1D$ which is $A_2$. He continues in this fashion, indefinitely. The total length of his path if $AB=5$ and $BC=12$ is of the form $a + b\sqrt{c}$. Find $\displaystyle\frac{abc}{4}$.
Consider infinite sequences $\{x_n\}$ of positive reals such that $x_0=1$ and $x_0\ge x_1\ge x_2\ge\ldots$.
[b]a)[/b] Prove that for every such sequence there is an $n\ge1$ such that: \[ {x_0^2\over x_1}+{x_1^2\over x_2}+\ldots+{x_{n-1}^2\over x_n}\ge3.999. \]
[b]b)[/b] Find such a sequence such that for all $n$: \[ {x_0^2\over x_1}+{x_1^2\over x_2}+\ldots+{x_{n-1}^2\over x_n}<4. \]
Professor Piraldo takes part in soccer matches with a lot of goals and judges a match in his own peculiar way. A match with score of $m$ goals to $n$ goals, $m\geq n$, is [i]tough[/i] when $m\leq f(n)$, where $f(n)$ is defined by $f(0) = 0$ and, for $n \geq 1$, $f(n) = 2n-f(r)+r$, where $r$ is the largest integer such that $r < n$ and $f(r) \leq n$.
Let $\phi ={1+\sqrt 5\over 2}$. Prove that a match with score of $m$ goals to $n$, $m\geq n$, is tough if $m\leq \phi n$ and is not tough if $m \geq \phi n+1$.
A sequence $ (a_1,b_1)$, $ (a_2,b_2)$, $ (a_3,b_3)$, $ \ldots$ of points in the coordinate plane satisfies \[ (a_{n \plus{} 1}, b_{n \plus{} 1}) \equal{} (\sqrt {3}a_n \minus{} b_n, \sqrt {3}b_n \plus{} a_n)\hspace{3ex}\text{for}\hspace{3ex} n \equal{} 1,2,3,\ldots.\] Suppose that $ (a_{100},b_{100}) \equal{} (2,4)$. What is $ a_1 \plus{} b_1$?
$ \textbf{(A)}\\minus{} \frac {1}{2^{97}} \qquad
\textbf{(B)}\\minus{} \frac {1}{2^{99}} \qquad
\textbf{(C)}\ 0 \qquad
\textbf{(D)}\ \frac {1}{2^{98}} \qquad
\textbf{(E)}\ \frac {1}{2^{96}}$
The sum to infinity of the terms of an infinite geometric progression is $ 6$. The sum of the first two terms is $ 4\frac {1}{2}$. The first term of the progression is:
$ \textbf{(A)}\ 3 \text{ or } 1\frac {1}{2} \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2\frac {1}{2} \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 9 \text{ or } 3$
Natural numbers $1, 2, 3,.., 100$ are contained in the union of $N$ geometric progressions (not necessarily with integer denominations). Prove that $N \ge 31$
Prove that the natural numbers can be divided into two groups in a way that both conditions are fulfilled:
1) For every prime number $p$ and every natural number $n$, the numbers $p^n,p^{n+1}$ and $p^{n+2}$ do not have the same colour.
2) There does not exist an infinite geometric sequence of natural numbers of the same colour.
Let a geometric progression with $n$ terms have first term one, common ratio $r$ and sum $s$, where $r$ and $s$ are not zero. The sum of the geometric progression formed by replacing each term of the original progression by its reciprocal is
$\textbf{(A) }\frac{1}{s}\qquad\textbf{(B) }\frac{1}{r^ns}\qquad\textbf{(C) }\frac{s}{r^{n-1}}\qquad\textbf{(D) }\frac{r^n}{s}\qquad \textbf{(E) }\frac{r^{n-1}}{s}$
Gizmo is thinking of a geometric sequence in which the third term is $1215$ and the fifth is $540$. Which of the following could be the eighth term of Gizmo's sequence?
$\text{(A) }-160\qquad\text{(B) }-135.5\qquad\text{(C) }216\qquad\text{(D) }240\qquad\text{(E) }472.5$