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 T O P I C    R E V I E W corchos68 Posted - 10/10/2012 : 14:59:19 A vendor at the fair sells an item for \$5. Every item after that is less 0.50 cents. How many items must you buy before you get a free one? First one is \$5, 2nd is \$4.50, 3rd is \$4 and so on until the eleventh item is free. Is there a way to write this as an algebra expression? 5   L A T E S T    R E P L I E S    (Newest First) the_hill1962 Posted - 11/26/2012 : 13:02:33 Looking at the title for this topic, the explanations are great.However, since this is in the "basic math" section, here is an simple explanation:5-0.5(11-1) = 0A lot of students get confused as to why "1" is subtracted.The easy way is to just refer to the original problem where it states "the FIRST one is \$5"Note that 5-0.5(1) = \$4.50, not \$5 as it wants.So, since "first" usually means n=1, you have to have 5-0.5(1-1) = \$5The 'simple explanation' for this problem would be to list the prices:1st is 5-0.5(1-1) = 52nd is 5-0.5(2-1) = 4.53rd is 5-0.5(3-1) = 44th is 5-0.5(4-1) = 3.55th is 5-0.5(5-1) = 36th is 5-0.5(6-1) = 2.57th is 5-0.5(7-1) = 28th is 5-0.5(8-1) = 1.59th is 5-0.5(9-1) = 110th is 5-0.5(10-1) = 0.511th is 5-0.5(11-1) = 0 Subhotosh Khan Posted - 11/20/2012 : 10:44:12 quote:Originally posted by corchos68A vendor at the fair sells an item for \$5. Every item after that is less 0.50 cents. How many items must you buy before you get a free one? First one is \$5, 2nd is \$4.50, 3rd is \$4 and so on until the eleventh item is free. Is there a way to write this as an algebra expression?a = 5an = an-1 - 0.5oran = 5 - 0.5 * (n-1) royhaas Posted - 10/12/2012 : 08:05:00 The clue lies in the fact that the difference between successive terms is constant. Perhaps conducting an Internet search for "arithmetic progression" or "arithmetic series" will help. corchos68 Posted - 10/11/2012 : 14:30:57 quote:Originally posted by royhaasDo you know what an arithmetic progression is?I don't. The homework asked if I could think of a different way to express my work - and I could not though I imagined there exists a simple or at least more elegant way than what I did. royhaas Posted - 10/11/2012 : 08:22:20 Do you know what an arithmetic progression is?

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