# SOLUTION: a. 1, 3, 5, 7, 9 b. 0, 50, 100, 150, 200 c. 3, 6, 12, 24, 48 d. 10, 100, 1,000, 10,000, 100,000 e. 9, 13, 17, 21, 25, 29 f. 1, 8, 27, 64, 125 Find the 100th term and the nt

Algebra ->  -> SOLUTION: a. 1, 3, 5, 7, 9 b. 0, 50, 100, 150, 200 c. 3, 6, 12, 24, 48 d. 10, 100, 1,000, 10,000, 100,000 e. 9, 13, 17, 21, 25, 29 f. 1, 8, 27, 64, 125 Find the 100th term and the nt      Log On

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 Question 336715: a. 1, 3, 5, 7, 9 b. 0, 50, 100, 150, 200 c. 3, 6, 12, 24, 48 d. 10, 100, 1,000, 10,000, 100,000 e. 9, 13, 17, 21, 25, 29 f. 1, 8, 27, 64, 125 Find the 100th term and the nth term for each of the sequences in exercise 2.Answer by Edwin McCravy(9719)   (Show Source): You can put this solution on YOUR website!``` a. 1, 3, 5, 7, 9 That's an arithmetic sequence with first term and common difference , because you always ADD 2 to get the next term. So we use ------------------------------------ b. 0, 50, 100, 150, 200 That's an arithmetic sequence with first term and common difference , because you always ADD 50 to get the next term. That one is done exactly like problem a. I'll let you do that one by yourself. ------------------------------------ c. 3, 6, 12, 24, 48 That's a geometric sequence with first term and common ration , because you always MULTIPLY BY 2 to get the next term. So we use ------------------------------------ d. 10, 100, 1,000, 10,000, 100,000 That's a geometric sequence with first term and common ratio , because you always MULTIPLY BY 10 to get the next term. That one is done exactly like problem c. I'll let you do that one by yourself. ------------------------------------ e. 9, 13, 17, 21, 25, 29 That's an arithmetic sequence with first term and , because you always ADD 4 to get the next term. That one is done exactly like problem a. I'll let you do that one by yourself. ------------------------------------ f. 1, 8, 27, 64, 125 Now that one is different from the others because it is neither an arithmetic sequence nor a geometric sequence. However we can recognize it as the sequence of the cube of the successive positive integers. 1=1³, 8=2³, 27=3³, 64=4³, 125=5³ So ------------------------------------ Edwin```