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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

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