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Module 9 Lesson 2 Common Sequences We saw in lesson one that numbers written in a specific order is called a sequence. There are two special sequences that we will look at in detail. Arithmetic sequences In a sequence, if the difference between any two consecutive terms is a constant, then the sequence is arithmetic. This difference, denoted d, is called the common difference. So π = π’2 − π’1 For example in the sequence 1, 4, 7, 10, … the common difference is 3 because 4-1 =3 and 7-4 = 3 and 10-7 = 3. To find a specific term π’π ( or ππ‘β term) of an arithmetic series, we use the formula ππ = ππ + ( π − π)π π’π is the specific term π’1 is the first term π is the number of the term you are finding π is the common difference For example: In 1, 4, 7, 10, … find π’8 So, π’8 = 1 + (8 − 1)3 = 1 + 7 β 3 = 1 + 21 = 22 Example set #1 Find the common difference for the following sequences 1. 2,6,10,14,… 2. 17, 9, 1, … 3. x, x+y, x+2y, x+3y, … Find π’12 for the following sequences. 4. 8, 6, 4, … 5. -4, -11, -18, -25, … Arithmetic Series The sum of an arithmetic series can be found using the formula: π ππ = 2 (π’1 + π’π ) For example, find the sum of the first 8 terms of -1, 2, 5, 8, … We know that π = 8,and π’1 = −1 We need to find π’8 . Utilizing the previous formula, π’8 = −1 + ( 8 − 1)( 3) = −1 + 7 β 3 = −1 + 21 = 20 Now π8 = 8 2 (−1 + 20 ) = 4 ( 19 ) = 76 Exercise set #2 1. Find π12 for 12, 6, -6, 0,… 2. Find 3+ 8+ 13 + 18+ ….+ 38 Geometric Sequences The second special sequence is the geometric sequence in which the ratio between consecutive terms is constant. This ratio, r, is called the common π’ ratio and can be calculated by dividing the second term by the first, π = π’2 . 1 The ππ‘β term of a geometric sequence can be found by using the formula π’π = π’1 π π−1 π’π = nth term π’1 = first term π= common ratio The sum of n terms is ππ = π’1 (1−π π ) 1−π Example set #3: Given 16, 8, 4, 2, 1, ½, … Find a) π’8 b) π8 An infinite geometric series is a special type of geometric series. Let’s look at a sequence with a common ratio of ½. If we start at 2, then the sequence would be: 2, 1, ½, ¼, 1/8, 1/16, 1/32, 1/64, 1/128, 1/256, … The sequence would never end, however we would never reach zero. Each individual term is getting smaller and smaller, closer and closer to zero. If we now look at the series: 2+ 1+ ½+ ¼ + 1/8 + 1/16+1/32+1/64+ 1/128 + 1/256+… We might ask how could we find a sum of a series that never ends? Let’s look at partial sums of the series. 2+ 1=3 2+ 1+ ½=3.5 2+ 1+ ½+ ¼=3.75 2+ 1+ ½+ ¼ + 1/8=3.875 2+ 1+ ½+ ¼ + 1/8 + 1/16=3.9375 2+ 1+ ½+ ¼ + 1/8 + 1/16+1/32=3.96875 2+ 1+ ½+ ¼ + 1/8 + 1/16+1/32+1/64=3.984375 2+ 1+ ½+ ¼ + 1/8 + 1/16+1/32+1/64+ 1/128=3.9921875 2+ 1+ ½+ ¼ + 1/8 + 1/16+1/32+1/64+ 1/128 + 1/256=3.99609375 As you can see, just by adding the first 10 terms, the sum changes very slowly. What would happen if we added the first 100 terms? The sum would change so insignificantly that it would be essentially adding zero each time. When the consecutive terms of an infinite geometric series get smaller and smaller, the common ratio would have to be between -1 and 1 or |r| < 1. The formula for finding the sum of an infinite geometric series when |r|<1 is π’1 π∞ = 1−π So in our example, the sum would be calculated by Example set #4 Find the infinite sum, if it exists 1. 1 25 1 1 + 250 + 2500 + β― 2. 3+ 9+27+… 2 1−1/2 2 = 1/2 = 4. Answers for exercise set #1 1. d= 6-2 = 4 2. d = 9 – 17 = -8 3. d = x + y – x = y 4. π’12 = 8 + ( 12 − 1)( −2 ) = 8 + 11 (−2) = 8 + −22 = −14 5. π’12 = −4 + ( 12 − 1 )( −7) = −4 + 11 ( −7 ) = −4 + −77 = −81 Answers for exercise set #2 1. π12 = 12 2 (12 + 78) = 540 2. 164 Answers for exercise set #3 1 8−1 1. π’8 = 16 (2) 2. π8 = 18 ) 2 1 1− 2 16(1− = 1 7 1 = 16 (2) 16 (128) = 16/128 255 ) 256 1 2 16( = 32 ( 255/256) Answers for exercise set #4 1. 1 25 1 1− 10 2 =45 2. No sum since r=3 which is greater than 1.