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 November 6th, 2011, 01:34 PM #1 Newbie   Joined: Sep 2011 Posts: 16 Thanks: 0 Arithmetic Series Can anyone help on this please A 40 year old buliding programme for new house began in the year 1951 (year 1) and finished in 1990 (year 40) The number of houses built each year form an arithmetic sequence with first term a and common difference d Given that 2400 new houses were built in 1960 and 600 new houses were built in 1990, find the value of a) the value of d b) the value of a c) the total number of houses built over the 40 year period Thankss
 November 6th, 2011, 02:10 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,900 Thanks: 1093 Math Focus: Elementary mathematics and beyond Re: Arithmetic Series Formula for nth term of arithmetic sequence with first term a and common difference d: a_n = a_1 + (n - 1)d. We have (1) 2400 = a + 9d (2) 600 = a + 39d Subtract (1) from (2): -1800 = 30d, d = -60, sub -60 for d in (1) to find a = 2940.

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### given that 2400 new houses were built in 1960 and 600 new houses were built in 1990, find

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