SOLUTION: An estate developer had a forty year building plan for a country starting from January 2001. The number of houses to be built each year forms an arithmetic progression (A.P). 372 h
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Question 1209824: An estate developer had a forty year building plan for a country starting from January 2001. The number of houses to be built each year forms an arithmetic progression (A.P). 372 houses were built in 2010 and plans to build 1032 houses in 2040. Find the:
(a) number of houses expected to be built in 2023;
(b) total number of houses expected from the developer at the end of forty years.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's solve this problem step-by-step.
**(a) Number of Houses Expected in 2023**
1. **Determine the Number of Terms:**
* 2010 is the 10th year (2010 - 2001 + 1 = 10).
* 2040 is the 40th year.
* 2023 is the 23rd year.
2. **Use the Arithmetic Progression (AP) Formula:**
* Let 'a' be the first term (number of houses in 2001).
* Let 'd' be the common difference.
* The nth term of an AP is given by: Tn = a + (n - 1)d
3. **Set Up Equations:**
* T10 = a + 9d = 372 ...(1)
* T40 = a + 39d = 1032 ...(2)
4. **Solve for 'a' and 'd':**
* Subtract equation (1) from equation (2):
* 30d = 660
* d = 22
* Substitute d = 22 into equation (1):
* a + 9(22) = 372
* a + 198 = 372
* a = 174
5. **Calculate the Number of Houses in 2023:**
* T23 = a + 22d = 174 + 22(22) = 174 + 484 = 658
* **Answer (a):** The number of houses expected to be built in 2023 is 658.
**(b) Total Number of Houses Expected at the End of Forty Years**
1. **Use the Sum of an AP Formula:**
* The sum of the first 'n' terms of an AP is given by: Sn = (n/2) * (a + Tn)
* We want to find S40.
2. **Calculate S40:**
* S40 = (40/2) * (a + T40)
* S40 = 20 * (174 + 1032)
* S40 = 20 * 1206
* S40 = 24120
* **Answer (b):** The total number of houses expected from the developer at the end of forty years is 24,120.
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