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.
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
i. The value of 'd':
To find the common difference 'd', we can use the information given. The number of houses built in 1960 is 2400, and the number of houses built in 1990 is 600. The difference between these two terms is 2400 - 600 = 1800. Since there are 30 years between 1960 and 1990, we divide the difference by 30 to find the common difference 'd'. Therefore, d = 1800/30 = 60.
ii. The value of 'a':
To find the first term 'a', we can use the information given. The number of houses built in 1960 is 2400, and the common difference 'd' is 60. We can use the formula for the nth term of an arithmetic progression: a + (n-1)d. Plugging in the values, we have: 2400 = a + (10-1)60. Simplifying this equation gives us: 2400 = a + 540. Subtracting 540 from both sides, we get: 1860 = a. Therefore, the value of 'a' is 1860.
iii. The total number of houses built over the 40-year period:
To find the total number of houses built, we can use the formula for the sum of an arithmetic progression: Sn = (n/2)(2a + (n-1)d). Plugging in the values, we have: S40 = (40/2)(2(1860) + (40-1)60). Simplifying this equation gives us: S40 = 20(3720 + 2340). Further simplification gives us: S40 = 20(6060). Multiplying 20 by 6060 gives us: S40 = 121,200. Therefore, the total number of houses built over the 40-year period is 121,200.
Let's solve this step-by-step:
i. To find the value of 'd', we need to use the formula for the nth term of an arithmetic progression (A.P.), which is given by:
a + (n-1)d
We are given that 2400 new houses were built in 1960, which means it is the 10th term of the A.P. (since 1960 - 1951 = 9 years have passed).
So, substituting the values into the formula, we have:
a + (10-1)d = 2400
Simplifying:
a + 9d = 2400
ii. To find the value of 'a', we can use the same formula with the information given for the 40th term:
a + (40-1)d = 600
Simplifying:
a + 39d = 600
iii. The total number of houses built over the 40-year period can be found by using the formula for the sum of an A.P., which is given by:
S = (n/2)[2a + (n-1)d]
In this case, n = 40 (since it's a 40-year program), and we need to find the value of 'a' and 'd' to substitute into the formula.
Let's solve for 'd' first:
From equation i: a + 9d = 2400 (equation i)
From equation ii: a + 39d = 600 (equation ii)
Subtracting equation i from ii, we have:
a + 39d - (a +9d) = 600 - 2400
30d = -1800
d = -1800/30
d = -60
Now, substituting the value of 'd' into equation i, we have:
a + 9(-60) = 2400
a - 540 = 2400
a = 2400 + 540
a = 2940
So, the value of 'd' is -60, and the value of 'a' is 2940.
Now, let's find the total number of houses built over the 40-year period using the formula for the sum of an A.P.:
S = (n/2)[2a + (n-1)d]
S = (40/2)[2(2940) + (40-1)(-60)]
S = 20[5880 - 2340]
S = 20(3540)
S = 70,800
Therefore, the total number of houses built over the 40-year period is 70,800.
To find the value of 'd' and 'a' in the arithmetic progression (A.P.), we will use the information given about the number of houses built in two specific years. We are given that 2400 new houses were built in 1960 (year 10) and 600 new houses were built in 1990 (year 40).
i. Finding the common difference 'd':
In an arithmetic progression, the common difference 'd' represents the difference between consecutive terms. We can find 'd' by subtracting the number of houses built in the earlier year from the number built in the later year and dividing it by the difference in the year numbers. Using the given information:
d = (number of houses in 1990 - number of houses in 1960) / (year number of 1990 - year number of 1960)
= (600 - 2400) / (40 - 10)
= -1800 / 30
= -60
Therefore, the value of 'd' is -60.
ii. Finding the first term 'a':
To find 'a', we need to determine the number of houses built in the first year (1951, year 1). Since we are given the value for year 10, we can use the formula for the nth term of an arithmetic progression to find 'a'. The formula is:
a + (n - 1) * d = nth term
Substituting the values into the formula:
a + (10 - 1) * -60 = 2400
a - 540 = 2400
a = 2400 + 540
a = 2940
Therefore, the value of 'a' is 2940.
iii. Finding the total number of houses built over the 40-year period:
To find the total number of houses built over the 40-year period, we can use the formula for the sum of the first 'n' terms of an arithmetic progression, which is:
Sum = (n/2) * (2a + (n - 1) * d)
Substituting the values into the formula:
Sum = (40/2) * (2 * 2940 + (40 - 1) * (-60))
Sum = 20 * (5880 - 39 * 60)
Sum = 20 * (5880 - 2340)
Sum = 20 * 3540
Sum = 70800
Therefore, the total number of houses built over the 40-year period is 70800.