1057 11.2 Arithmetic Sequences and Series Now find d. an = a1 + 1n - 12d Formula for the nth term a17 = a1 + 117 - 12d Let n = 17. -13 = 35 + 16d Let a17 = -13 and a1 = 35; subtract. -48 = 16d Subtract 35. d = -3 Divide by 16 and interchange sides. S Now Try Exercise 61. Any sum of the form an i=1 1 di +c2, where d and c are real numbers, represents the sum of the terms of an arithmetic sequence having first term a1 = d112 + c = d + c and common difference d. These sums can be evaluated using the formulas in this section. The TI-84 Plus will give the sum of a sequence without having to first store the sequence. The screen here illustrates this method for the sequences in Example 9. EXAMPLE 9 Using Summation Notation Evaluate each sum. (a) a10 i=1 14i + 82 (b) a 9 k=3 14 - 3k2 SOLUTION (a) This sum contains the first 10 terms of the arithmetic sequence having a1 = 4 # 1 + 8 = 12, First term and a10 = 4 # 10 + 8 = 48. Last term Using the formula Sn = n 2 1a1 + an2, we obtain the following sum. a10 i=1 14i + 82 = S10 = 10 2 112 + 482 = 51602 = 300 (b) The first few terms are 34 - 31324 + 34 - 31424 + 34 - 31524 + g = -5 + 1-82 + 1-112 + g . Thus, a1 = -5 and d = -3. If the sequence started with k = 1, there would be nine terms. Because it starts at 3, two of those terms are missing, so there are seven terms and n = 7. Use the formula Sn = n 232a1 + 1n - 12d4. a9 k=3 14 - 3k2 = 7 2 321-52 + 17 - 121-324 = -98 S Now Try Exercises 69 and 71.
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