The problem describes an arithmetic progression (AP) of houses built each year starting from 2001. We are given that 372 houses were built in 2010 and 1032 houses are planned to be built in 2040. We need to find: (a) the number of houses expected to be built in 2023, and (b) the total number of houses expected from the developer at the end of the forty years.

1. Problem Description

The problem describes an arithmetic progression (AP) of houses built each year starting from

2

0

0

1. We are given that 372 houses were built in 2010 and 1032 houses are planned to be built in

2

0

4

0. We need to find:

(a) the number of houses expected to be built in 2023, and

(b) the total number of houses expected from the developer at the end of the forty years.

2. Solution Steps

Let $a_n$ be the number of houses built in the $n$-th year after 2000, so $a_1$ is for 2001, $a_2$ for 2002, etc. The sequence $a_n$ forms an arithmetic progression.

We are given that $a_{10} = 372$ (houses built in 2010) and $a_{40} = 1032$ (houses built in 2040).

The formula for the $n$-th term of an arithmetic progression is:

$a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference.

Using the given information:

$a_{10} = a_1 + (10-1)d = a_1 + 9d = 372$ (1)

$a_{40} = a_1 + (40-1)d = a_1 + 39d = 1032$ (2)

Subtract equation (1) from equation (2):

$(a_1 + 39d) - (a_1 + 9d) = 1032 - 372$

$30d = 660$

$d = \frac{660}{30} = 22$

Now, substitute $d = 22$ into equation (1):

$a_1 + 9(22) = 372$

$a_1 + 198 = 372$

$a_1 = 372 - 198 = 174$

So, the first term $a_1 = 174$ and the common difference $d = 22$.

(a) To find the number of houses expected to be built in 2023, we need to find $a_{23}$:

$a_{23} = a_1 + (23-1)d = 174 + 22(22) = 174 + 484 = 658$

Therefore, the number of houses expected to be built in 2023 is

6

5

8.

(b) To find the total number of houses expected from the developer at the end of the forty years, we need to find the sum of the arithmetic progression from $a_1$ to $a_{40}$, which is $S_{40}$. The formula for the sum of an arithmetic progression is:

$S_n = \frac{n}{2}(a_1 + a_n)$

$S_{40} = \frac{40}{2}(a_1 + a_{40}) = \frac{40}{2}(174 + 1032) = 20(1206) = 24120$

Therefore, the total number of houses expected from the developer at the end of the forty years is

2

4

1

2

0.

3. Final Answer

(a) The number of houses expected to be built in 2023 is

6

5

8. (b) The total number of houses expected from the developer at the end of the forty years is

2

4

1

2

0.