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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: 20,966 Thanks: 2216 (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
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Re: Arithmetic progression

Quote:
 Originally Posted by noki 2) In arithmetic progression 28,26,24,... find a term that is equal with the fifth part of the sum of preliminary terms.
You mean: In arithmetic progression 28,26,24,... find the first term
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 Tags arithmetic, progression Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post will.i.am1 Algebra 14 May 29th, 2013 06:27 AM Daksh Algebra 3 November 22nd, 2012 09:08 PM Trafford99 Algebra 7 February 28th, 2012 03:07 PM panky Algebra 4 November 17th, 2011 07:05 AM Francis410 Algebra 1 March 22nd, 2011 07:02 AM

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