Lesson 12–3 Objectives Be able to find the terms of an ARITHMETIC sequence Be able to find the sums of arithmetic series

Arithmetic sequence – when the terms differ by the same number This is called the common difference, “d”

Example 1: Identifying Arithmetic Sequences Determine whether the sequence could be arithmetic. If so, find the common first difference and the next term. –10, –4, 2, 8, 14, … –10, –4, 2, 8, 14 Differences 6 6 6 6 The sequence could be arithmetic with a common difference of 6. The next term is 14 + 6 = 20.

Example 1: Identifying Arithmetic Sequences Determine whether the sequence could be arithmetic. If so, find the common first difference and the next term. –2, –5, –11, –20, –32, … –2, –5, –11, –20, –32 Differences –3 –6 –9 –12 The sequence is not arithmetic because the first differences are not common.

Example 1: Identifying Arithmetic Sequences Determine whether the sequence could be arithmetic. If so, find the common first difference and the next term. 1.9, 1.2, 0.5, –0.2, –0.9, ... 1.9, 1.2, 0.5, –0.2, –0.9 –0.7 Differences The sequence could be arithmetic with a common difference of –0.7. The next term would be –0.9 – 0.7 = –1.6.

Formulas for arithmetic sequences: Recursive formula: an = an–1 + d Explicit formula: an = a1 + (n – 1)d

Example 2: Finding the nth Term Given an Arithmetic Sequence Find the 12th term of the arithmetic sequence 20, 14, 8, 2, 4, .... Step 1 Find the common difference: d = 14 – 20 = –6.

 Step 2 Evaluate by using the formula. an = a1 + (n – 1)d General rule. Substitute 20 for a1, 12 for n, and –6 for d. a12 = 20 + (12 – 1)(–6) = –46 The 12th term is –46. Check Continue the sequence. 

Example 2: Finding the nth Term Given an Arithmetic Sequence Find the 11th term of the arithmetic sequence. –3, –5, –7, –9, … Step 1 Find the common difference: d = –5 – (–3)= –2. Step 2 Evaluate by using the formula. an = a1 + (n – 1)d General rule. Substitute –3 for a1, 11 for n, and –2 for d. a11= –3 + (11 – 1)(–2) = –23 The 11th term is –23.

Example 2: Finding the nth Term Given an Arithmetic Sequence Find the 11th term of the arithmetic sequence. 9.2, 9.15, 9.1, 9.05, … Step 1 Find the common difference: d = 9.15 – 9.2 = –0.05. Step 2 Evaluate by using the formula. an = a1 + (n – 1)d General rule. Substitute 9.2 for a1, 11 for n, and –0.05 for d. a11= 9.2 + (11 – 1)(–0.05) = 8.7 The 11th term is 8.7.

Example 3: Finding Missing Terms Find the missing terms in the arithmetic sequence 17, , , , –7. Step 1 Find the common difference. an = a1 + (n – 1)d General rule. Substitute –7 for an, 17 for a1, and 5 for n. –7 = 17 + (5 – 1)(d) –6 = d Solve for d.

Step 2 Find the missing terms using d= –6 and a1 = 17. = 11 The missing terms are 11, 5, and –1. a3 = 17 +(3 – 1)(–6) = 5 a4 = 17 + (4 – 1)(–6) = –1

Example 3: Finding Missing Terms Find the missing terms in the arithmetic sequence 2, , , , 0. Step 1 Find the common difference. an = a1 + (n – 1)d General rule. 0 = 2 + (5 – 1)d Substitute 0 for an, 2 for a1, and 5 for n. –2 = 4d Solve for d.

Step 2 Find the missing terms using d= and a1= 2. The missing terms are = 1

You can use the explicit formula even if you don’t know the first term…

Example 4: Finding the nth Term Given Two Terms Find the 5th term of the arithmetic sequence with a8 = 85 and a14 = 157. Step 1 Find the common difference. Let an = a14 and a1 = a8. Replace 1 with 8. a14 = a8 + (14 – 8)d a14 = a8 + 6d Simplify. Substitute 157 for a14 and 85 for a8. 157 = 85 + 6d 72 = 6d 12 = d

Step 2 Write a rule for the sequence, and evaluate to find a5. Repeat the process again, this time letting “a5” be the unknown a14 = a5 + (14 – 5)d 157 = a5 + (14 – 5)12 157 = a5 + 108 49 = a5 The 5th term is 49.

Example 4: Finding the nth Term Given Two Terms Find the 11th term of the arithmetic sequence. a2 = –133 and a3 = –121 Step 1 Find the common difference. an = a1 + (n – 1)d a3 = a2 + (3 – 2)d Let an = a3 and a1 = a2. Replace 1 with 2. a3 = a2 + d Simplify. –121 = –133 + d Substitute –121 for a3 and –133 for a2. d = 12

Step 2 Write a rule for the sequence, and evaluate to find a11. Repeat the process again, this time letting “a11” be the unknown a3 = a11 + (3 – 11)d –121 = a11 + (3 – 11)12 –121 = a5 – 96 –25 = a11 The 11th term is -25.

Example 4: Finding the nth Term Given Two Terms Find the 11th term of each arithmetic sequence. a3 = 20.5 and a8 = 13 Step 1 Find the common difference. an = a1 + (n – 1)d General rule Let an = a8 and a1 = a3. Replace 1 with 3. a8 = a3 + (8 – 3)d a8 = a3 + 5d Simplify. 13 = 20.5 + 5d Substitute 13 for a8 and 20.5 for a3. –7.5 = 5d Simplify. –1.5 = d

Step 2 Write a rule for the sequence, and evaluate to find a11. Repeat the process again, this time letting “a11” be the unknown a8 = a11 + (8 – 11)d 13 = a11 + (8 – 11)(–1.5) 13 = a5 + 4.5 8.5 = a11 The 11th term is 8.5.

Arithmetic series – the sum of specified terms in an arithmetic sequence

Example 5: Finding the Sum of an Arithmetic Series Find the indicated sum for the arithmetic series. S18 for 13 + 2 + (–9) + (–20) + ... Find the common difference. d = 2 – 13 = –11 Find the 18th term. a18 = 13 + (18 – 1)(–11) = –174

 Sum formula Substitute. = 18(-80.5) = –1449 Check Use a graphing calculator. 

Example 5: Finding the Sum of an Arithmetic Series Find the indicated sum for the arithmetic series. Find S15. Find 1st and 15th terms. a1 = 5 + 2(1) = 7 a15 = 5 + 2(15) = 35 = 15(21) = 315

Check Use a graphing calculator. 

Example 5: Finding the Sum of an Arithmetic Series Find the indicated sum for the arithmetic series. S16 for 12 + 7 + 2 +(–3)+ … Find the common difference. d = 7 – 12 = –5 Find the 16th term. a16 = 12 + (16 – 1)(–5) = –63

Find S16. Sum formula. Substitute. = 16(–25.5) Simplify. = –408

Example 5: Finding the Sum of an Arithmetic Series Find the indicated sum for the arithmetic series. Find 1st and 15th terms. a1 = 50 – 20(1) = 30 a15 = 50 – 20(15) = –250

Find S15. Sum formula. Substitute. = 15(–110) Simplify. = –1650

Example 6: Theater Application The center section of a concert hall has 15 seats in the first row and 2 additional seats in each subsequent row. How many seats are in the 20th row? Write a general rule using a1 = 15 and d = 2. an = a1 + (n – 1)d Explicit rule for nth term a20 = 15 + (20 – 1)(2) Substitute. = 15 + 38 Simplify. = 53 There are 53 seats in the 20th row.

How many seats in total are in the first 20 rows? Find S20 using the formula for finding the sum of the first n terms. Formula for first n terms Substitute. Simplify. There are 680 seats in rows 1 through 20.

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