ARITHMETIC SEQUENCE, MEAN AND SERIES WEEK 2 QUARTER 1

ARITHMETIC SEQUENCE, MEAN AND SERIES WEEK 2 QUARTER 1

• 1.
  WELCOME Students!
• 2.
  DRILL Subtraction of Integers
• 3.
  (-5) - (-2)
• 4.
  (8) - (-3)
• 5.
  (-9) - (8)
• 6.
  (4) - (10)
• 7.
  (-3) - (11)
• 8.
  WHO AM I?
• 9.
  FINITE OR INFINITE? ARITHMETIC OR NOT?
• 10.
  82, 86, 90,94, 98
• 11.
  2, 4, 6,8, 10, 12, ...
• 12.
  10, 20, 30,40, ....
• 13.
  5, 25, 125,625
• 14.
  Quarter 1 Week 2
• 15.
  a. Determine thenth term of a given arithmetic sequence; b. Define an arithmetic mean; c. Find the arithmetic mean between the terms of an arithmetic sequence; d. Describe an arithmetic series; e. Determine the sum of the first n terms of an arithmetic sequence and; f. Confidently solve different problems involving arithmetic sequence, arithmetic mean, and arithmetic series. O B J E C T I V E S
• 16.
  Lesson 1 DETERMINING the Nthof an ARITHMETIC SEQUENCE
• 17.
  ARITHMETIC SEQUENCE is a sequencewhere every term after the first term is obtained by adding a constant.
• 18.
  common difference is theconstant number added to the preceding term of the arithmetic sequence. It can be calculated by subtracting any 2 consecutive term in the arithmetic sequence
• 19.
  ARITHMETIC SEQUENCE FORMULA: an = a1+ ( n - 1 )d
• 20.
  FORMULA: an = a1+ ( n - 1 )d nth term first term number of terms common difference
• 21.
  3, 7, 11,15,19 3 7 11 15 19 = 3 = 3+4 4 4 4 4 = 3+4+4 = 3+0(4) = 3+1(4) = 3+2(4) = 3+3(4) =3+4(4) = a1 = a1+0d = 3+4+4+4 = 3+4+4+4+4 = a2 = a1+1d = a3 = a1+2d = a4 = a1+3d = a5 = a1+4d d an = a1+(n-1)d
• 22.
  ARITHMETIC SEQUENCE FORMULA: an = a1+ ( n - 1 )d
• 23.
  Find the 16thterm of the arithmetic sequence 15, 21, 27, 33, 39, … Problem 1
• 24.
  Step 1: Identifythe given and the unknown in the problem. an = a1 + ( n - 1 )d Problem: Find the 16th term of the arithmetic sequence 15, 21, 27, 33, 39, … an = ? a1 n d = 15 = 16 = 6
• 25.
  Step 2: Substitutethe known quantities to the formula. an = a1 + ( n - 1 )d a16 = 15 + ( 16 - 1 )6 a1 n d = 15 = 16 = 6
• 26.
  Step 3: Simplify a16= 15 + ( 16 - 1 ) = 15 + ( 15 )6 = 15 + 90 a16 = 105 a16 = 15 + ( 16 - 1 )6 = 15 + ( 15 )6 = 15 + 90 Therefore, the 16th term is 105.
• 27.
  In the arithmetic sequence 4,10, 16, 22, 28,… which term is 130? Problem 2
• 28.
  Step 1: Identifythe given and the unknown in the problem. an = a1 + ( n - 1 )d an = 130 a1 n d = 4 = ? = 6 Problem: In the arithmetic sequence 4, 10, 16, 22, 28,… which term is 112?
• 29.
  Step 2: Substitutethe known quantities to the formula. an = a1 + ( n - 1 )d 130 = 4 + ( n - 1 )6 an a1 d = 130 = 4 = 6
• 30.
  Step 3: Simplify. 130= 4 + ( n - 1 )6 130 = 4 + 6n - 6 130 = -2 + 6n
• 31.
  Step 3: Simplify. 130= -2 + 6n 130 + 2 = 6n 132 = 6n 22 = n Therefore, 130 is the 22nd term.
• 32.
  Find the 16thterm of the arithmetic sequence whose first term is 11 and whose seventh term is 59. Problem 3
• 33.
  Step 1: Identifythe given and the unknown in the problem. a1 = 11 a7 = 59 n =7 d= ? Find the 16th term of the arithmetic sequence whose first term is 11 and whose seventh term is 59. a16 =?
• 34.
  Step 2: Solvefor d an = a1 + ( n - 1 )d Find the 16th term of the arithmetic sequence whose first term is 11 and whose seventh term is 59. a7 = a1 + ( n - 1 )d 59 = 11 + ( 7 - 1 )d
• 35.
  Step 2: Solvefor d Find the 16th term of the arithmetic sequence whose first term is 11 and whose seventh term is 59. 59 = 11 + ( 6 )d 59 - 11 = 6d 48 = 6d 8 = d
• 36.
  shortcut in finding d
• 37.
  a1 = 11a7 = 59 Find the 16th term of the arithmetic sequence whose first term is 11 and whose seventh term is 59. 2. 7 - 1 = 6 1. 59 - 11 = 48 3. 48 ÷ 6 = 8 d
• 38.
  Step 1: Identifythe given and the unknown in the problem. a1 = 11 a7 = 59 n =7 d= 8 Find the 16th term of the arithmetic sequence whose first term is 11 and whose seventh term is 59. a16 =?
• 39.
  Step 2: Solvefor a16 an = a1 + ( n - 1 )d Find the 16th term of the arithmetic sequence whose first term is 11 and whose seventh term is 59. a16 = a1 + ( n - 1 )d a16 = 11 + ( 16 - 1 )8
• 40.
  Step 2: Solvefor a16 Find the 16th term of the arithmetic sequence whose first term is 11 and whose seventh term is 59. a16 = 11 + ( 16 - 1 )8 = 11 + ( 15 )8 = 11 + 120 a16 = 131 Therefore, the 16th term is 131.
• 41.
  Lesson 2 Arithmetic Mean
• 42.
  Arithmetic Means - arethe terms between any two nonconsecutive terms of an arithmetic sequence Steps in solving the arithmetic mean 2. Using the formula of arithmetic sequence solve for the value of the common difference 3. To get the value of the missing term add the common difference and the value before the missing term. 1. Identify the no. of terms in the arithmetic sequence.
• 43.
  Problem 1: What isthe arithmetic mean between 50 and 120?
• 44.
  formula: a + b 2 Solution a= 50 b = 120 a + b = 50+120 2 2 = 170 2 = 85 AM =
• 45.
  Problem 2: Insert 6arithmetic mean between 5 and 61.
• 46.
  Insert six arithmeticmeans between 5 and 61. 5, Given: 2 1 4 3 6 5 8 7 61, ___, ___, ___, ___, ___, ___, Formula: Solution: 13, 21, 29, 37, 45, 53, a1= 5 a8= 61 n= 8 a2 = a1 + d Solve for d. 1. 61 - 5 = 56 2. 8 - 1 = 7 3. 56 ÷ 7 = 8 = d = 5 + 8 a2 = 13 a3 = a2 + d a3 = 13 + 8 = 21 d= ?
• 47.
  Problem 3: If the3rd term of an AP is18 and the 9th term is 60, what is the first term and the 6th term of the sequence?
• 48.
  __, __,18,__,__,__,__,__,60 a3 a9 a4a5 a6 a7 a8 a1 a2 a1 =18 a7 =60 1. 60 – 18 = 42 Solve for d. 2. 7 – 1 = 6 3. 42/6 = 7 a2 = 18 – 7 = 11 a1 = 11 – 7 = 4 a4 = 18 + 7 = 25 a5 = 25 + 7 = 32 a6 = 32 + 7 = 39 4 39 a3 = 18
• 49.
  Lesson 3 Arithmetic Series
• 50.
  Arithmetic Series - Itis the sum of the terms of a given Arithmetic Sequence.
• 51.
  Formula: Sn = n( a1 + an ) 2 Sn – sum of n terms an – nth terms a1 – first terms n – no. of terms
• 52.
  Formula: Sn = n[2a1 + (n-1)d] 2 Sn – sum of n terms an – nth terms a1 – first terms n – no. of terms d – common difference
• 53.
  Find the sumof the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ... Problem 1
• 54.
  Step 1 :Identify the known and unknown quantities. S20 = ? n = 20 a1 = 18 Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ...
• 55.
  Formula #1 Sn =n ( a1 + an ) 2 S20=? n=20 a1=18 an=? Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ...
• 56.
  Step 2: Solvefor a20 an = a1 + ( n - 1 )d a20 = a1 + ( n - 1 )d a20 = 18 + ( 20 - 1 )5 Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ...
• 57.
  a20 = 18+ ( 20 - 1 )5 = 18 + ( 19 )5 = 18 + 95 a20 = 113 Therefore, the 20th term is 113. Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ...
• 58.
  Sn = n( a1 + an ) 2 S20 = ? n = 20 a1 = 18 a20 = 113 Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ...
• 59.
  S = n( a1 + an ) 2 Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ... S20 = 20 ( 18 + 113 ) 2
• 60.
  Find the sumof the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ... S20 = 20 ( 18 + 170 ) 2 S20 = 10 ( 131 ) S20 = 1 310 )
• 61.
  Formula #2: Sn =n [2a1 + (n-1)d] 2 Sn – sum of n terms an – nth terms a1 – first terms n – no. of terms d – common difference
• 62.
  Step 1 :Identify the known and unknown quantities. Find the sum of the first twenty terms of the arithmetic sequence 18, 23, 28, 33, 38, ... S20=? n=20 a1=18 d = 5 an=?
• 63.
  Formula #2: Sn =n [2a1 + (n-1)d] 2 S20=? n=20 a1=18 d = 5 an=? S20 = 20 [2(18) + (20-1)5] 2
• 64.
  S20 = 20[2(18) + (20-1)5] 2 S20 = 10 [36 + (19)5] S20 = 10 [36 + 95] S20 = 10 [ 131 ] S20 = 1 310
• 65.
  Thank you!