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April 5th, 2012, 01:36 PM  #1 
Guest Joined: Posts: n/a Thanks:  Arithmetic progression
1) How many numbers have between 100 and 500? a) even b) odd 2) In arithmetic progression 28,26,24,... find a term that is equal with the fifth part of the sum of preliminary terms. 
April 5th, 2012, 02:16 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038 
(1a). See (1b). The answer will depend on whether 100 and 500 are included. (1b). There are 200 odd numbers between 100 and 500, given by 99 + 2k, where k is any integer from 1 to 200. (2). This question could be clearer, but 20 = (28 + 26 + 24 + 22)/5. 
April 5th, 2012, 09:46 PM  #3  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,290 Thanks: 1023  Re: Arithmetic progression Quote:
that is equal to 1/5 of the sum of the preceeding terms; right? You can do this by formula (combining arithmetic series formulas): n = term number a = first term d = common difference This would be a general case formula for "1/5 preceeding" scenario: n = [13d  2a + SQRT(4a^2  4ad + 121d^2)] / (2d) Substitute a=28 and d=2 and you'll get n = 5 or n = 36 n = 5 is the solution Skipjack gave you: 5th term = 20 n = 36 is also valid: 36th term = 42 ; terms 1 to 35 add to 210  

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