# SOLUTION: Can someone please help me with this problem? Find a1 in a geometric series for which Sn=189, r=1/2, and an=3.

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 Algebra: Sequences of numbers, series and how to sum them Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Sequences-and-series Question 77523This question is from textbook Algebra 2 : Can someone please help me with this problem? Find a1 in a geometric series for which Sn=189, r=1/2, and an=3.This question is from textbook Algebra 2 Answer by Edwin McCravy(8909)   (Show Source): You can put this solution on YOUR website!``` Can someone please help me with this problem? Find a1 in a geometric series for which Sn=189, r=1/2, and an=3. an = a1rn-1 an = a1(1/2)n-1 3 = a1(1/2n-1) 3(2n-1) = a1 a1(1 - rn) Sn = ------------ 1 - r Sn = a1(1/2)n-1 a1[1 - (1/2)n] 189 = ---------------- 1 - 1/2 Simplifying the second a1[1 - 1/2n] 189 = -------------- 1/2 a1(1 - 2-n) 189 = ------------- 1/2 Multiply the numerator and denominator on the right by 2 189 = 2a1(1 - 2-n) Since 3(2n-1) = a1, substitute 3(2n-1) for a1 189 = 2[3(2n-1)](1 - 2-n) 189 = 3(21)(2n-1)(1 - 2-n) 189 = 3(2n)(1 - 2-n) Divide both sides by 3 63 = 2n(1 - 2-n) Distribute 63 = 2n - 2n-n 63 = 2n - 20 63 = 2n - 1 64 = 2n 26 = 2n 6 = n 3(2n-1) = a1 3(26-1) = a1 3(25) = a1 3(32) = a1 96 = a1 To check, start with 96 and multiply by 1/2 until we have 6 terms: 96, 48, 24, 12, 6, 3 So the nth or 6th term, an = a6 = 3. That checks. Sn = S6 = 96+48+24+12+6+3 = 189. That checks. Edwin```