1960 ---> term10 = a + 9d = 2400
1990 ---> term40 = a + 39d = 600
subtract them
30d = -1800
d = -60
then a + 9(-60) = 2400
a = 2940
sum40 = 20(2(2940) + 39(-60))
= 70800
a 40 year building programme for new houses began in 1951(year 1) and finished in 1990 year (40). The number of houses built each year form an A.P. With the first term 'a' and the common difference 'd'. Given that 2400 new houses were built in 1960 and 600 new houses werw built in 1990, find:
i. The value of 'd'
ii. The value of 'a'
iii. The total number of houses built over the 40 year period.
1 answer