# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2009 | June | Q#5

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Question

A 40-year building programme for new houses began in Oldtown in the year 1951 (Year 1) and finished in 1990 (Year 40).

The numbers 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

a.  the value of d,

b.  the value of a,

c.  the total number of houses built in Oldtown over the 40-year period.

Solution

a.

We are given that point following data of Arithmetic Series;  We are given that number of houses built in 1961 (10th Year) is 2400. Therefore; We are given that number of houses built in 1990 (40th Year) is 600. Therefore; Expression for the general term in the Arithmetic Progression (A.P) is: Therefore, for both 10th and 40th terms we can write equations as;

 For 10th term; For 40th term;    We can solve these two simultaneous equations to fin the values of ‘d’.  Substituting this value of ‘a’ in equation of 40th term;      b.

We have found two equations in terms of a and d in (a).  We can simultaneously solve these two equations to find ‘a’.

From (a) we have found that; Substituting this value of ‘d’ in equation of 40th term;     c.

We are required to find the total number of houses built in Oldtown over the 40-year period.

Since numbers of houses built each year form an arithmetic sequence the sum of all houses built is  the sum of this arithmetic sequence.

Expression for the sum of number of terms in the Arithmetic Progression (A.P) is: Therefore;      