1. ALGEBRA II HONORS ARITHMETIC and GEOMETRIC SERIES

2. SERIES : The sum of an arithmetic or geometric sequence. Put your writing instruments down and watch Find S 100 for the arithmetic series 7 + 13 + 19 + 25 + 31 + … Let’s find t 100.t n = t 1 + (n – 1)d t 100 = 7 + (100 – 1)(6) = 601 Therefore, t 100 = 601, t 99 = 595, t 98 = 589, t 97 = 583, t 96 = 577, etc…

3. So, now we have 7 + 13 + 19 + 25 + 31 + … + 577 + 583 + 589 + 595 + 601 Just for grins, let’s pair up the terms of the series. A) What do you notice about each pair of numbers? SOLUTION : They all add up to 608. B) How many pairs of numbers are there? SOLUTION : 50

4. C) So, what do we have so far? SOLUTION : S 100 = (50)(608) = 30,400 D) How did we get the 50 from the original information? SOLUTION : E) Where does the 608 first come from? SOLUTION : 7 + 601 or t 1 + t 100 So, a formula that works for this problem is

5. So, an arithmetic series formula is Since t n = t 1 + (n – 1)d, we can substitute to get : So, a second arithmetic series formula is Note the red type above. Ah, hem, aren’t you supposed to be doing something now?

6. Given the geometric series 7 + 21 + 63 + …, find S 100. You may put your down. S 100 = 7 + 73 1 + 73 2 + 73 3 + 73 4 + … + 73 99 F) Now, multiply both sides of the equation by 3. Hmm, why 3? Subtract. - 3S 100 = 73 1 + 73 2 + 73 3 + 73 4 + … + 73 99 + 73 100 S 100 – 3S 100 = 7 - 7 3 100 S 100 (1 – 3) = 7(1 – 3 100 ) Factor Divide by (1 – 3)

7. So, now we have G) What is the formula for a geometric series? GEOMETRIC SERIES FORMULA S n = t 1 1 – r n 1 - r Psst, note the color above. Don’t tell anyone.

8. 1) Find the 127 th partial sum of the arithmetic series with t 1 = 17 and d = 4. SOLUTION : Which formula will be the best one to use? We know the first term and the common difference. So use : = 34,163 Therefore, the sum of the first 127 terms is 34,163.

9. 2) Find S 34 for the geometric series with t 1 = 7 and r = 1.03. SOLUTION : = 404.1112356 Therefore, the sum of the first 34 terms of the series is 404.1112356.

10. 3) 30705 is a partial sum in the arithmetic series with first term 17 and common difference 3. Which partial sum is it? SOLUTION : Which formula will we use? Solving gives you the solutions 138 and. But, a term number can not be negative, so the only answer is 138.

11. 4) 50238.14 is the approximate value of a partial sum in the geometric series with t 1 = 150 and r = 1.04. Which term is it? SOLUTION : (÷ 150) (simplify denom.) (cross x) (x -1) Yikes! n = 67.9999 or 68 Therefore, there are 68 terms in the series. Remember a term number must be a whole number. (-1)

12. 5) SOLUTION : Is this an arithmetic or a geometric series? It is arithmetic because the variable is not an exponent. If you are not sure, find the first 3 terms of the series and find the difference or ratio. Because it’s easier! n = 20, t 1 = 3(1)-5 = -2, and t 20 = 3(20)-5 = 55 = 530

13. 6) SOLUTION : What type of series is this? Geometric, since the variable is an exponent. n = 20 since 0-19 inclusive is 20 t 1 = 2 0 = 1, t 2 = 2 1 = 2 = 1,048,575

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