#arithmetic sequences

[gender-blendered]

Thanks to my math teacher who allowed me to sperg out about my beautiful boy on my infinite arithmetic sequence assignment

[karolldocs]

Watch "INTRODUCTION TO ARITHMETIC SERIES + DERIVATION OF FORMULAS (PART ONE)" on YouTube

NEW VIDEO ABOUT ARITHMETIC SERIES. ❤❤

[davidrussellschilling]

Arithmetic Sequence

C O N T E N T S:

GENERAL INFO

KEY TOPICS

It is possible for a list of numbers to appear to be an arithmetic sequence based on the first few terms, but then fail after that.(More…)

Note that in an arithmetic sequence, you always add (or subtract) the same number to get from one term to the next.(More…)

POSSIBLY USEFUL

The formula for finding any term in the sequence is Un (or Ur) the 1st term +…

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[infiniteeducation]

Help with ARITHMETIC SEQUENCES AND SERIES.

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[precal30]

Arithmetic Sequences

An arithmetic sequence is when a sequence of real numbers have a constant distance between each consecutive term, where a consecutive term is when two numbers are beside each other and the constant is called the common difference, d. An infinite arithmetic sequence is when the sequence continues on into positive or negative infinity, normally indicated by three periods. Arithmetic sequence: 1, 2, 3, 4 Infinite arithmetic sequence: 1, 2, 3, 4, ... An arithmetic sequence can either be increasing or decreasing by a constant term, d. If the arithmetic sequence is increasing, the common difference, d, is positive. If the arithmetic sequence is decreasing, the common difference, d, is negative. Common Difference, d d = t₂ - t₁ For the following infinite arithmetic sequence: 4, 7, 10, 13, ... The common difference, d, is the following: d = t₂ - t₁ d = 7 - 4 d = 3 If we found the common difference, d, for other consecutive terms, it would still equal 3. This concludes that the sequence is arithmetic. When graphing an arithmetic sequence, it will always be in linear shape, with the points (n, tn). Arithmetic Sequence, tn tn = t₁ + d(n - 1) tn = general term (mostly the "imaginary" last term of the sequence), said as "nth term" t₁ = first term d = common difference n = number of terms in the sequence Finding a Term Value in a Sequence 1. Using the arithmetic sequence formula, set up the formula with respect to the term needed to be found. 2. Find the common difference, d, using the arithmetic sequence and common difference formula, and find the first term, t₁, by using the initial term in the arithmetic sequence. 2.1. If the common difference, d, and the first term, t₁, are unable to be found, construct and simplify a system of linear equations with the two terms that are given, where the largest term goes on top of the linear system. 2.1.1. Subtract them from each other, where the largest term goes on top of the linear system, eliminating the first term, t₁, enabling you to solve for the common difference, d. 2.1.2. Use any one of the two linear equations from the system to substitute the common difference, d, in, and solve for the first term, t₁. 3. With all variables found, substitute them into the arithmetic sequence formula and solve for the missing term value. Finding the Number of Terms in a Sequence 1. Using the arithmetic sequence formula, set up the formula with respect to the term value given, imagining that that term is the "last term", tn. 2. Find the common difference, d, using the arithmetic sequence and common difference formula, and find the first term, t₁, by using the initial term in the arithmetic sequence. 3. With all variables found, substitute them into the arithmetic sequence formula and factor the common difference, d, and the expression (n - 1), and solve for the number of terms, n, in the arithmetic sequence if tn was the last term. If n = a, then tn = tₐ.

PRACTICE QUESTIONS (BOOKLET) Page: 8 #4, 6, 10, 13, 15

Given the following infinite arithmetic sequence: -3, 2, 7, 12, ... Determine t20. 1. Using the arithmetic sequence formula, set up the formula with respect to the term needed to be found. Since the term needed to be found is t20, we will set up the formula with respect to t20. Setting up the formula, where tn = t20, n = 20: tn = t₁ + d(n - 1) t20 = t₁ + d(20 - 1) t20 = t₁ + d(19) 2. Find the common difference, d, using the arithmetic sequence and common difference formula, and find the first term, t₁, by using the initial term in the arithmetic sequence. Finding the common difference, d, where t₂ = 2, t₁ = -3: d = t₂ - t₁ d = 2 - (-3) d = 2 + 3 d = 5 Finding the first term, t₁, by looking at the arithmetic sequence and seeing the initial term: t₁ = -3 3. With all variables found, substitute them into the arithmetic sequence formula and solve for the missing term value. Substituting values, where d = 5, t₁ = -3: t20 = t₁ + d(19) t20 = (-3) + (5)(19) Multiplying 5 and 19: t20 = (-3) + (5)(19) t20 = (-3) + 95 Adding -3 and 95: t20 = (-3) + 95 t20 = 92 Therefore, the 20th term, t20, is 92, or t20 = 92.

Given the following information: t₃ = 81 and t12 = 27 Determine t23. 1. Using the arithmetic sequence formula, set up the formula with respect to the term needed to be found. Since the term needed to be found is t23, we will set up the formula with respect to t23. Setting up the formula, where tn = t23, n = 23: tn = t₁ + d(n - 1) t23 = t₁ + d(23 - 1) t23 = t₁ + d(22) 2. Find the common difference, d, using the arithmetic sequence and common difference formula, and find the first term, t₁, by using the initial term in the arithmetic sequence. There is no way of finding the common difference, d, or the first term, t₁, with the given information, so we skip to step 2.1. 2.1. If the common difference, d, and the first term, t₁, are unable to be found, construct and simplify a system of linear equations with the two terms that are given. Setting up the formula, where tn = t₃ = 81, n = 81: tn = t₁ + d(n - 1) t₃ = t₁ + d(3 - 1) 81 = t₁ + d(2) 81 = t₁ + 2d Setting up the formula, where tn = t12 = 27, n = 27: tn = t₁ + d(n - 1) t12 = t₁ + d(12 - 1) 27 = t₁ + d(11) 27 = t₁ + 11d 2.1.1. Subtract them from each other, where the largest term goes on top of the linear system, eliminating the first term, t₁, enabling you to solve for the common difference, d. The largest term is t12, so its linear equation will go on top. 27 = t₁ + 11d 81 = t₁ + 2d Subtracting the linear system: 27 = t₁ + 11d 81 = t₁ + 2d -54 = 0 + 9d -54 = 9d Solving for the common difference, d, by dividing both sides by 9: -54 = 9d -6 = d d = -6 2.1.2. Use any one of the two linear equations from the system to substitute the common difference, d, in, and solve for the first term, t₁. Using the first linear equation, t12's, where d = -6, and solving for t₁: 27 = t₁ + 11d 27 = t₁ + 11(-6) Multiplying 11 and -6: 27 = t₁ + 11(-6) 27 = t₁ + (-66) Adding 66 to both sides: 27 = t₁ + (-66) 93 = t₁ t₁ = 93 3. With all variables found, substitute them into the arithmetic sequence formula and solve for the missing term value. Substituting values, where d = -6, t₁ = 93: t23 = t₁ + d(22) t23 = 93 + (-6)(22) Multiplying -6 and 22: t23 = 93 + (-6)(22) t23 = 93 + -132 Adding 93 and -132: t23 = 93 + -132 t23 = -39 Therefore, the 23rd term, t23, is -39, or t23 = -39.

Given the following infinite arithmetic sequence: -3, 2, 7, 12, ... What term in the infinite arithmetic sequence has the value of 212? 1. Using the arithmetic sequence formula, set up the formula with respect to the term value given, imagining that that term is the "last term", tn. Let us imagine that 212 is the value of the last term, tn, therefore making tn = 212. Setting up the formula, where tn = 212: tn = t₁ + d(n - 1) 212 = t₁ + d(n - 1) 2. Find the common difference, d, using the arithmetic sequence and common difference formula, and find the first term, t₁, by using the initial term in the arithmetic sequence. Finding the common difference, d, where t₂ = 2, t₁ = -3: d = t₂ - t₁ d = 2 - (-3) d = 2 + 3 d = 5 Finding the first term, t₁, by looking at the arithmetic sequence and seeing the initial term: t₁ = -3 3. With all variables found, substitute them into the arithmetic sequence formula and factor the common difference, d, and the expression (n - 1), and solve for n. If n = a, then tn = tₐ. Substituting values, where d = 5, t₁ = -3: 212 = t₁ + d(n - 1) 212 = (-3) + (5)(n - 1) Factoring 5 and (n - 1): 212 = (-3) + (5)(n - 1) 212 = (-3) + 5n - 5 Adding -3 and -5: 212 = (-3) + 5n - 5 212 = 5n - 8 Adding 8 to both sides: 212 = 5n - 8 220 = 5n Dividing both sides by 5: 220 = 5n 44 = n n = 44 Therefore, there are 44 terms in the arithmetic sequence if the last term was 212. This means, 212 is the 44th term, and so t44 = 212.

[infiniteeducation]

For more help with ARITHMETIC SEQUENCE SUMMATION:

https://www.youtube.com/watch?v=xoHyki4S_So

[infiniteeducation]

For more help with arithmetic sequences: http://www.onlinemathlearning.com/arithmetic-sequences-algebra.html

[maths-help]

could you help? the first three terms of an arithmetic sequence are: 2k + 3, 5k - 2 and 20k - 15. a) show that k= 4, b) find the value of the first three terms of the sequence, c) write down the value of the common difference, d) calculate the 20th term of sequence. so sorry its this much i am desperate for help, since i have IB exams on monday..

Okay, there might be a language problem here but as I understand it, arithmetic sequence means that the difference between two consecutive numbers is constant.

2*4 + 3 = 115*4 - 2 = 1820*4 - 15 = 65

Since 7 is not equal to 47, I don’t see how we can show that k=4?

In my mind, k=2/3!

Here’s how I figured it out (this should also solve b) and c) !):

I hope it’s readable!

The value of the of the 20th term of the sequence is

13/3 + (-3)*19 = -158/3

I hope this helps with your exams a little <3

-Frauke