Arithmetic Sequence PDF

Arithmetic Sequence LESSON PROPER An arithmetic sequence is a sequence in which consecutive terms have a common difference. The next term of the sequence is obtained by adding a constant number from the preceding term or previous term. The sequence a1, a2, a3, …, an, is arithmetic...

Arithmetic Sequence LESSON PROPER An arithmetic sequence is a sequence in which consecutive terms have a common difference. The next term of the sequence is obtained by adding a constant number from the preceding term or previous term. The sequence a1, a2, a3, …, an, is arithmetic if there is a number d such that: a2 – a1 = d, a3 – a2 = d, a4 – a3 = d and so on. The number d is the common difference in the arithmetic sequence. Formula for the nth term of an arithmetic sequence The nth term, an, of an arithmetic sequence with first term, a1, and common difference, d, is given by an = a1 + (n – 1) d EXAMPLE 1.1 nth term of an arithmetic sequence Write a formula for the nth term of the given arithmetic sequence. a. 11, 18, 25, 32, 39, … Solution: a. Each term is 7 more than the previous term, so d = 7. Since a 1 = 11, the nth term is: an = a1 + (n – 1) d Use the general formula for the nth term of arithmetic sequence. = 11 +(n – 1)7 Substitute the given values. = 11 + 7n – 7 Perform distribution method. an = 7n + 4 Simplify to get the desired formula Thus, the formula for the nth term of the given arithmetic sequence is an = 7n + 4. To check if the formula is correct, we can Substitute few terms from the given sequence to the formula we obtain. Learning continues in the new normal …Page 1 of 15  a1 = 7(1) + 4 = 11 As you can see the given terms suited the a2 = 7(2) + 4 = 18 given formula that’s why our formula is a3 = 7(3) + 4 = 25 correct EXAMPLE 1.2 nth term of an arithmetic sequence If there is an indicated number of terms asked, we can use this process to find that term: Find the 15th term of the arithmetic sequence 16, 20, 24, 28, 32, … Solution: Note that a1 = 16, d = 20 – 16 = 4, and n = 15 an = a1 + (n – 1) d Use the general formula for the nth term of arithmetic sequence. a15 = 16 + (15 – 1) 4 Substitute the given values. a15 = 16 + (14) 4 Simplify to find the final answer, multiply an = 7n + 4 14 to 4. a15 = 72 The 15th term of the given sequence is 72. Note: If the other parts of the arithmetic sequence were asked, we can derive the general formula with respect to the part being asked. EXAMPLE 1.3 nth term of an arithmetic sequence On this example the missing value is n (position of the term on the sequence), for this question we can use the following process as long as we have the given. In the arithmetic sequence 6, 12, 18, 24, 30, …, which term is 120? Solution: Find n when an = 120 Let a1 = 6, d = 6, and an = 120 Learning continues in the new normal …Page 2 of 15  an = a1 + (n – 1) d Use the general formula for the nth term of arithmetic sequence. 120 = 6 + (n – 1) 6 Substitute the given values. 120 = 6 + 6n – 6 Apply the distributive property. 120 = 6n Combine like terms. 20 = n Divide both sides by 6, because we must cancel out the number on the side with n Thus, 120 is the 20th term EXAMPLE 1.4 nth term of an arithmetic sequence On this part before we obtain the missing nth term, we will find first the common difference using only an indicted term on the sequence. Find the 18th term of the arithmetic sequence which first term is 11 and the seventh term is 59. Solution: Since the given only are two nonconsecutive terms of the sequence we must find first the common difference. Find the common difference, d, by substituting a1 = 11 and a7 = 59 into the formula an = a1 + (n – 1) d. a7 = a1 + (7 – 1) d Use the formula for the nth term of arithmetic sequence substituting the value of n of the bigger term. 59 = 11 + (7 – 1) d Replace a7 with 59 and a1 with 11. 59 = 11 + 6d Simplify. 59 – 11 = 6d Transpose 11 to the other side to combine like terms. 48 = 6d Simplify. 8=d Divide both sides by 6 to get the common difference. Learning continues in the new normal …Page 3 of 15  Since we find the common difference of the sequence, we can find the term being asked on the question. Find a18 by substituting a1 = 11 and d= 8 in an = a1 + (n – 1) d a18 = 11 + (18 – 1) 8 = 11 + (17) 8 Therefore our 18th term is equal to 147 = 11 + 136 = 147 Real life application of finding the nth term of an arithmetic sequence A conference hall has 20 rows of seats. The first row contains 20 seats, the second row contains 22 seats, the third row contains 24 seats, and so on. How many seats are there in the last row? Solution: We know that n = 20, a1 = 20, a2 = 22, a3 = 24, and d= 2 an = a1 + (n – 1) d Use the general formula for the nth term of arithmetic sequence. a20 = 20 + (20 – 1) 2 Substitute the given values. = 20 + (19) 2 Simplify. = 20 + 38 = 58 The number of seats on the last row is equal to 58 Learning continues in the new normal …Page 4 of 15  Arithmetic Series Arithmetic Series is the indicated sum of an arithmetic sequence. Carl Friedrich Gauss – a German Mathematician who discovers the concept of arithmetic series in his elementary days by solving a problem involving adding the first 100 counting numbers in a short method. Given an arithmetic sequence the sum of the first n terms (arithmetic series) can be obtained using the following formula: 𝑛 𝑠𝑛 = (𝑎1 + 𝑎𝑛 ) Given the value of n, the first term 2 (a1) and the nth term (an) EXAMPLE Sum of the first nth term of the arithmetic sequence (Arithmetic Series) Find the sum of the following: First 25 terms of the arithmetic sequence 18, 23, 28, 33, 38, … a. Using the first formula First, find a25 with a1 = 18, d = 5, and n = 25 an = a1 + (n – 1) d Use the general formula for the nth term of arithmetic sequence. a25 = 18 + (25 – 1) 5 Substitute the given values. = 18 + (24) 5 Simplify. = 138 The 25th term is 138 Solve for S25: 𝑛 Sum of the first n terms of an arithmetic 𝑠𝑛 = (𝑎1 + 𝑎𝑛 ) 2 sequence. 25 𝑠25 = (18 + 138) Substitute the known values. 2 𝑠25 = 12.5 (156) Simplify 𝑠25 = 1950 Therefore, the sum of the first 25 terms of the arithmetic sequence is 1950. Learning continues in the new normal …Page 5 of 15 Real life application of finding the Arithmetic series. A conference hall has 20 rows of seats. The first row contains 20 seats, the second row contains 22 seats, the third row contains 24 seats, and so on. How many seats are there in the conference hall? We know that n = 20, a1 = 20, a2 = 22, a3 = 24, and d= 2 an = a1 + (n – 1) d Use the general formula for the nth term of arithmetic sequence. a20 = 20 + (20 – 1) 2 Substitute the given values. = 20 + (19) 2 Simplify. = 20 + 38 = 58 The number of seats on the last row is equal to 58 Solve for S20: 𝑛 Sum of the first n terms of an 𝑠𝑛 = (𝑎1 + 𝑎𝑛 ) 2 arithmetic sequence. 20 𝑠20 = (20 + 58) Substitute the known values. 2 𝑠20 = 10 (78) Simplify 𝑠20 = 780 Arithmetic Mean The terms of an arithmetic sequence that are between two given terms are called arithmetic means. Insert four arithmetic means between 14 and 144 Solution: Look for numbers m1, m2, m3, m4, such that 14, m1, m2, m3, m4, 44 is an arithmetic sequence. The first step to find the arithmetic mean is to find the common difference. Learning continues in the new normal …Page 6 of 15  an = a1 + (n – 1) d Use the general formula for the nth term of arithmetic sequence. a6 = a1 + (6 – 1) d We use 6 as the value of n because we have 44 = 14 + (6 – 1) d 6 terms. 44 = 14 + 5d 44 – 14 = 5d 30 = 5d By applying the step on finding the common difference the value of d is 6. 6=d Since d = 6, we have: m1 = a1 + d m 2 = m1 + d m3 = m 2 + d m4 = m3 + d = 14 + 6 = 20 + 6 = 26 + 6 = 32 + 6 = 20 = 26 = 32 = 38 Thus, the four-arithmetic means between 14 and 44 are 20, 26, 32, and 38 Learning continues in the new normal …Page 7 of 15  Geometric Sequence Geometric sequence is a sequence in which each term is obtained by multiplying the preceding term by a fixed or constant number called as the COMMON RATIO. Each geometric sequence can be expressed in terms of r and its previous term. The constant ratio is called a common ratio represented by r. Formula for the nth term of a Geometric Sequence an = a1  rn – 1 Where: an = nth term of the geometric sequence a1 = first term r = common ratio EXAMPLE nth term of a geometric sequence sequence If there is no desired formula and you asked to find a certain term of a geometric sequence just use the general form for finding the nth term. Find the 8th term of the geometric sequence 3, 6, 12, 24, 48,.... Solution: 6 a1 = 3 and r = =2 3 an = a1 rn – 1 Use the general formula for the nth term of a geometric sequence a8 = 3 (2)8 – 1 Substitute the given values to the general formula. a8 = 3 (2)7 Simplify. a8 = 3 (128) a8 = 384 Multiply, then find the final answer Thus, the 8th term of the given geometric sequence is equal to 384. Learning continues in the new normal …Page 8 of 15  Geometric Means and Series The indicated sum of a geometric sequence is called a GEOMETRIC SERIES. Sum of the first n terms of a Geometric Sequence The sum of a finite geometric sequence is calculated by the formula below: 𝑎1 (1− 𝑟 𝑛 ) 𝑠𝑛 = 1− 𝑟 Where Sn is the sum of first n terms of the sequence, a1 is the first term, r is the common ratio n is the indicated number of terms. A finite geometric sequence is a sequence with the indicated number of terms, it has an end. Infinite geometric sequence a sequence with a continuous number of terms, it has no end, or the end cannot be determined. Geometric Series Find the sum of the following: a. First twelve terms of the geometric sequence 3, - 9, 27, - 81, 243, … Solution: a. Given a1 = 3 r=-9÷3=-3 Solve for S12 by substituting n with 12, a1 with 3, and r with – 3 into the formula. 𝑎1 (1− 𝑟 𝑛 ) 𝑠𝑛 = Use the formula for sum of 1− 𝑟 the terms a finite geometric sequence Learning continues in the new normal …Page 9 of 15  3[(1−(− 3)12 )] 𝑠12 = Substitute the known 1−( −3 ) values of the given sequence 3(1− 531441) 𝑠12 = Perform the indicated 1+3 operation and simplify 3(− 531440) 𝑠12 = 4 𝑠12 = - 398 580 The final answer is - 398 580 Geometric Means Insert two geometric means between 7 and 875 Solution: There are four terms in the given sequence: 7, ___, ___, 875 First, find the common ratio using the formula a4 = a1 r 4 – 1 Use the general formula for the nth term of a geometric sequence 875 = 7r3 Substitute the known values 125 = r3 Simplify 5=r the common ratio of the given sequence is equal to 5 To complete the sequence, multiply 7 by 5, and then multiply the result by 5 also. 75, 35, 175, 875 Therefore, the required geometric means are 35 and 175 Learning continues in the new normal …Page 10 of 15  Harmonic and Fibonacci Sequence HARMONIC SEQUENCE If a , a , …, a are terms of an arithmetic sequence, then the sequence of 1 2 n 1 1 1 1 reciprocals of these terms 𝑎 , 𝑎 , 𝑎 , …. 𝑎 is called harmonic sequence. 1 2 3 𝑛 EXAMPLE nth term of Harmonic Sequence 1 1 1 Find the seventh term of the harmonic sequence 2, 6, 10 an = a1 + (n – 1) d Use the general formula for the nth term of arithmetic sequence to solve for the nth term of harmonic sequence. = 2 +(7 – 1)4 Substitute the given values. = 2 + (6)4 Perform the indicated operation = 2 + 24 Simplify to get the answer = 26 Get the reciprocal of the answer to convert it to harmonic sequence. The 1 Final answer is 26 HARMONIC MEAN If a and b are two positive numbers, then the harmonic mean of a and b is given by: 2𝑎𝑏 𝐻= a+𝑏 EXAMPLE: Find the harmonic mean of the following a. 60 and 40 → a = 60 b = 40 2(60)(40) 𝐻= 60+40 4800 𝐻= 100 𝐻 = 48 Learning continues in the new normal …Page 11 of 15 FIBONACCI SEQUENCE Introduced by Leonardo Fibonacci also known as “Leonardo of Pisa”. In 1202, Fibonacci investigated a problem involving the speed of reproduction of rabbits. The Fibonacci Sequence consist of the following numbers: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,… Learning continues in the new normal …Page 12 of 15 Learning continues in the new normal …Page 13 of 15 Learning continues in the new normal …Page 14 of 15 Learning continues in the new normal …Page 15 of 15