Question 1209824
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.