Sequences and Series -- Arithmetic and Geometric Sequences

Sequences are fun to learn and not really that difficult. 
There are many similarities between arithmetic and geometric sequences, so 
learn both together. 

Enjoy !!!!! 

From Mthcounts Mini: Sequences and Series

Easier concepts:

Sequences

Arithmetic sequence/determine the nth term

Arithmetic and geometric sequences

Mathcounts strategies : review some sums 

Note : Don't just memorize, but really understand the concepts.

Harder concepts:

Sum and Average of An Evenly Space

Relationship between arithmetic sequences, mean and median

Sequences, series and patterns

Some Common Sums

A Skill for the 21st Century: Problem Solving by Richard Rusczyk

Does our approach to teaching math fail even the smartest kids ? 

Quotes from that article  "According to research from the University of California, Los Angeles, as many as 60 percent of all college students who intend to study a STEM (science, technology, engineering, math) subject end up transferring out. In an era when politicians and educators are beside themselves with worry over American students’ lagging math and science scores compared to the whiz kids of Shanghai and Japan, this attrition trend so troubles experts it has spawned an entire field of research on “STEM drop-out,” citing reasons from gender and race to GPAs and peer relationships."

Why Science Majors Change Their Minds (It's Just So Darn Hard) from the New York Times 

A Skill for the 21st Century: Problem Solving by Richard Rusczyk, founder of  "Art of Problem Solving".

Top 10 Skills We Wish Were Taught at School, But Usually Aren't 
from Lifehacker

Som sums' formula:

V's record

 V's record 

weekly homework about 30 to 45 minutes 

extra videos, links optional 

First lesson:

2010 chapter sprint: 

Hi, I got a score of 22 and got questions 16, 22, 23, 24,25,27,29, and 30 wrong. 

I just guessed these questions because I didn't really find a way to do them.

Second meet: 

2011-12 Mathcounts handbook  (40 questions total) 

Warm Up 1: 8
Warm Up 2: 19,20

Warm Up 3: 32, 39 

Workout 1: 23, 24, 30

Third meet: 

2010 Mathcounts school test : 

Hi, I finished trying the Sprint and Target questions:

 these are the problems that I got wrong 

Sprint: I got 7/15 correct.

     15,16: Attempted but answer was wrong

     19,20,26,27,29: Didn't attempt

       30: Couldn't find a good method to do, but was able to solve it by listing out all the possibilities.

Target: I got 6/8 correct.

    The first 6 were relatively easy and I could find a clear way to do them

    The last 2, I couldn't find a way to approach the problem.

Fourth meet : 

2011 Mathcounts school test : last 15 sprint and last 4 target 

Sprint: Out of the last 15, 

I got 7 correct.

The questions that I got wrong were 20,21,22,24,27,28,29,30. 

I tried 21,22, and 27 but the answers were incorrect. I wasn't able to attempt the rest. 

Target: I got every question other than #5. 

However, questions 1 and 2, 

I got wrong at first, but when I retried them, 

I was able to get them. I read the problems wrong and didn't fully understand them the first time.

Fifth meet : review 

Sixth meet : 

2011 chapter 

Hi Mrs Lin! I was able to try both the target and the sprint round questions and here are my results:

Target: I missed 3 & 8, and I didn't attempt them.

Sprint: I got 17,21,23,24,25, 27,29,30.

I didn't attempt 27,29, and 30,

2012 Mathcounts school 

Hi Mrs Lin! I was able to finish both the chapter and target, and here are my results:

Target: I got 5,6,and 8 wrong. I didn't try any of them because I didn't know how to do them.

Sprint: I got 18,23,25,26, and 30

Apr. 30, 2023 

2012 Mathcounts chapter 

Hi Mrs. Lin! I hope your having a very good day and week! I tried both the target and the sprint. 

Target: I got only number 8 wrong, but I didn't know how to do it.

Sprint: I got number 22, 24, 26,28,29, and 30 wrong. I tried to do 22, but I got it wrong. The rest I didn't know how to do them..

May 7th, review 

May 13th, 2023 

2013 Mathcounts School 

Hi Mrs.Lin, I was able to try both the Target and the Sprint and here are my results!

Target (5/8) I got 6,7, and 8 wrong, however, I was able to figure out the answer to problem 6 when I reviewed it. 

Sprint: I got 30,29,28, and 24 wrong. 

May 20th, 2023

Hi Mrs Lin!

2013 Mathcounts Chapter 

I was able to try both the target and sprint round tests and here are my results:

Target: 6/8. I attempted the first 6 problems and got all of them right. I understood the problems fairly well and was able to do all of them on the first try.

sprint: I got 10/15 right. I got all problems I attempted right, and didn't attempt 30,28,25,24,21, and 19.

2014 Mathcounts school 

Hi Mrs Lin, I was able to finish both the target and the sprint, and here were my results!

Sprint: 22/30, I didn't get 18,19,24,25,26,28,29, and 30. I attempted number 18, but didn't get it.

Target: I got 5 out of 8 on the target, but after reviewing my answers, I was able to figure out number 5.

2015 Mathcounts school 

Hi Mrs Lin, I was able to try both the sprint and target. For the sprint, I got problems 20 and 27 wrong, and I didn't attempt any problems past 24 other than 27.

For my  Target I got numbers 6,7, and 8 wrong, and I didn't know how to approach any of them. 

Pathfinder

From Mathcounts Mini :

Counting/Paths Along a Grid

From Art of Problem Solving

Counting Paths on a Grid 

Math Principles : Paths on a Grid : Two Approaches 

Question #1: How many ways to move the dominoes on a 6 by 6 checker board if you can only move the dominoes to the right or to the bottom starting from the upper left and you can't move the dominoes diagonally? 

Solution :
You can move the dominoes 5 times to the right at most and 5 down to
the bottom at most, so the answer is \(\dfrac {\left( 5+5\right) !} {5! \times 5!}\) = 252 ways

Question # 2: How many ways can you  move from A to B if you can only move downward and to right? 

Solution : There are \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) * 2 * \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) = 9800 ways from A to B

Sum of All the Possible Arrangements of Some Numbers

Check out Mathcounts, the best middle school competition math program up to the national level.

Questions to ponder: (detailed solutions below) 

It's extremely important for you to spend some time pondering on these questions first without peeking on the solutions. 

#1: Camy made a list of every possible distinct four-digit positive integer that can be formed using each of the digits 1, 2 , 3 and 4 exactly once in each integer. What is the sum of the integers on Camy's list?

#2: Camy made a list of every possible distinct five-digit positive even integer that can be formed using each of the digits 1, 3, 4, 5 and 9 exactly once in each integer. What is the sum of the integers on Camy's list? (2004 Mathcounts Chapter Sprint #29)

#3: 2020 Mathcounts state sprint #24 













Solutions:
#1:  
Solution I: 
There are 4! = 24 ways to arrange the four digits. Since each digit appears evenly so each number will appear 24 / 4 = 6 times.
1 + 2 + 3 + 4 = 10 and 10 times 6 = 60 ; 60 (1000 +100 +10 + 1) = 60 x 1111 = 66660, which is the answer.

Solution II: 

The median of four numbers 1, 2, 3, 4 is (1 + 2 + 3 + 4) / 4 = 2.5 and there are 4! = 24 ways to arrange
the four numbers. 
2.5 (1000 + 100 + 10 + 1) x 24 = 66660 

#2:  
Solution I: 
Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely.
Again there are 4! = 24 ways to arrange the four numbers. 1 + 3 + 5 + 9 = 18 and 18 x 6 = 108 (Each number that can be moved freely appears 6 times evenly.)108 x 11110 + 4 x 24 = 1199976

Solution II: 

Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely. 
There will be 4! = 24 times the even number 4 will be used so 4 x 24 = 96
As for the remaining 4 numbers, their average (or mean) is \(\dfrac{1 + 3+ 5 + 9} {4} = 4.5\)                               
4.5 * ( 10000 + 1000 + 100 + 10) * 24 (arrangements)  + 96 = 4.5 * 11110 * 24 + 96 = 1199976

#3: The answer is 101. 

Other applicable problems: (answers below)

#1: What is the sum of all the four-digit positive integers that can be written with the digits 1, 2, 3, 4 if each digit must be used exactly once in each four-digit positive integer? (2003 Mathcounts Sprint #30)

#2: What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once? (You can use a calculator for this question.) (2005 AMC-10 B)

#3: What is the sum of all the four-digit positive integers that can be written with the digits 2, 4, 6, 8 if each digit must be used exactly once in each four-digit positive integer?

#4: What is the sum of all the 5-digit positive odd integers that can be written with the digits 2, 4, 6, 8, and 3 if each digit must be used exactly once in each five-digit positive integer?  

#5:What is the sum of all the four-digit positive integers that can be written with the digits 2, 3, 4, 5 if each digit must be used exactly once in each four-digit positive integer?

  





Answer key: 
#1: 66660
#2: \(\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8\)
4.8 * 11111 =\(\color{red}{53332.8}\) 
#3: 133320
#4: 1333272
#5: 93324

2015 Mathcounts State Prep: Mathcounts State Harder Questions

Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.

Question: 2010 Mathcounts State Team Round #10: A square and isosceles triangle of equal height are side-by-side, as shown, with both bases on the x-axis. The lower right vertex of the square and the lower left vertex of the triangle are at (10, 0). The side of the square and the base of the triangle on the x-axis each equal 10 units. A segment is drawn from the top left vertex of the square to the farthest vertex of the triangle, as shown. What is the area of the shaded region?

 There are lots of similar triangles for this question, but I think this is the fastest way to find the area.
  \(\Delta \)AGB is similar to \(\Delta \)DGC and their line ratio is 15 to 10 or 3 : 2.
     \(\Delta \)CGF is similar to \(\Delta \)CBE. 
     \(\dfrac {CG} {CB}=\dfrac {GF} {BE}\)
     \(\dfrac {2} {5}=\dfrac {GF} {10}\)\(\rightarrow GF=4\) From there you get the area =
      \(\dfrac{10\times 4} {2}=20\)

Question: 2010 Mathcounts State #30:  Point D lies on side AC of equilateral triangle ABC such that the measure of angle DBC is 45 degrees. What is the ratio of the area of triangle ADB to the area of triangle CDB? Express your answer as a common fraction in simplest radical form.

Since each side is the same for equilateral triangle ABC, once you use the 30-60-90 degree angle ratio and 45-45-90 degree angle ratio, you'll get the side.
Since area ratio stays constant, you can plug in any numbers and it's much easier to use integer first so I use 2 for \(\overline {CD}\).
From there you get the side length for each side is \(\sqrt {3}+1\).
\(\overline {AC}-C\overline {D}=\sqrt {3}+1-2\) = \(\overline {AD} = \sqrt {3}-1\)
\(\Delta ABD\) and \(\Delta CBD\) share the same vertex, so their area ratio is just the side ratio, which is \(\dfrac {\sqrt {3}-1} {2}\).