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November 6th, 2011, 12:34 PM   #1
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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

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November 6th, 2011, 01:10 PM   #2
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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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