5.5 Geometric Series (1/7) In an geometric series each term increases by a constant multiplier (r) This means the difference between consecutive terms is a common ratio (r) Therefore: r T11 ÷ T10 = T37 ÷ T36 = T317 ÷ T316 = T11 T10 = T37 T36 = T317 T316 = r OR

5.5 Geometric Series (2/7) T2 T1 T3 T2 m 7 847 m Example: Find m if 7 + m + 847 is a geometric series. Note: T2 T1 = T3 T2 m 7 = 847 m Therefore: m2 = 5 929 m = 77

5.5 Geometric Series (3/7) T2 T1 T3 T2 Example: Is 13 + 169 + 2028 + … a geometric series? Note: T2 T1 = T3 T2 = r Therefore: 169 13 2028 169 = 13 and = 12 13 12 Therefore NOT Geometric Series

5.5 Terms of a Geometric Series (4/7) The terms of an geometric series are: a + ar + ar2 + ar3 + … + arn-1 + … Tn = a r n – 1 Value of nth term Common Ratio Term number 1st term

5.5 Terms of a Geometric Series (5/7) Example: 3 4 6 12 12 36 Find the common ratio of + + + … 6 12 3 4 2 3 r = ÷ = Find the 100th term Tn = a r n – 1 ( ) 100-1 3 4 2 3 = x 3 x 299 22x399 297 398 = =

5.5 Terms of a Geometric Series (6/7) Example: 1 + 3 + 9 + … Which term in the series above is 243. a = 1 Tn = a rn – 1 r = 9 ÷ 3 = 3 243 = 1 x 3n – 1 Tn = 243 3n – 1 = 243 log33n – 1 = log3243 log 243 log 3 n - 1 = log3243 = n - 1 = 5 n = 6

5.5 Terms of a Geometric Series (7/7) The 3rd term of a series is 100 and the 5th term is 2500. Example: Find r. Find a. ar5-1 = ar4 = 2500 Tn = a rn – 1 ar3-1 = ar2 = 100 100 = a x 53 – 1 ar4 ar2 2500 100 = 100 = a x 52 100 = a x 25 r2 = 25 a = 4 r = 5