The sums of the 5th and 13th terms of an Arithmetic Progression are 21 and 37 respectively.
Determine:
i) The 40th term. (3 Marks)
ii) The sum of the 29th term. (3 Marks)

Let's solve the problem step by step.

### Given:
- The sum of the 5th and 13th terms of an Arithmetic Progression (AP) is 21.
- The sum of the 13th and 21st terms is 37.

### Step 1: Express the terms in terms of the first term \( a \) and the common difference \( d \).

- The general formula for the \( n \)-th term of an AP is given by:
  \[
  T_n = a + (n-1)d
  \]

- For the 5th term \( T_5 \):
  \[
  T_5 = a + 4d
  \]

- For the 13th term \( T_{13} \):
  \[
  T_{13} = a + 12d
  \]

- For the 21st term \( T_{21} \):
  \[
  T_{21} = a + 20d
  \]

### Step 2: Set up the equations based on the problem statement.

- The sum of the 5th and 13th terms:
  \[
  T_5 + T_{13} = (a + 4d) + (a + 12d) = 2a + 16d = 21 \quad \text{(Equation 1)}
  \]

- The sum of the 13th and 21st terms:
  \[
  T_{13} + T_{21} = (a + 12d) + (a + 20d) = 2a + 32d = 37 \quad \text{(Equation 2)}
  \]

### Step 3: Solve the system of equations.

1. **From Equation 1**:
   \[
   2a + 16d = 21 \quad \text{(Equation 1)}
   \]
   Divide by 2:
   \[
   a + 8d = 10.5 \quad \text{(Equation 3)}
   \]

2. **From Equation 2**:
   \[
   2a + 32d = 37 \quad \text{(Equation 2)}
   \]
   Divide by 2:
   \[
   a + 16d = 18.5 \quad \text{(Equation 4)}
   \]

### Step 4: Subtract Equation 3 from Equation 4.

\[
(a + 16d) - (a + 8d) = 18.5 - 10.5
\]
\[
8d = 8
\]
\[
d = 1
\]

### Step 5: Substitute \( d = 1 \) back into Equation 3 to find \( a \).

\[
a + 8(1) = 10.5
\]
\[
a + 8 = 10.5
\]
\[
a = 2.5
\]

### Step 6: Determine the required values.

i) **The 40th term** \( T_{40} \):

\[
T_{40} = a + 39d = 2.5 + 39(1) = 2.5 + 39 = 41.5
\]

ii) **The sum of the first 29 terms** \( S_{29} \):

The sum of the first \( n \) terms of an AP is given by:
\[
S_n = \frac{n}{2} \times (2a + (n-1)d)
\]

For \( n = 29 \):
\[
S_{29} = \frac{29}{2} \times (2(2.5) + 28(1)) = \frac{29}{2} \times (5 + 28) = \frac{29}{2} \times 33 = \frac{957}{2} = 478.5
\]

### Final Answers:
i) The 40th term is **41.5**.

ii) The sum of the first 29 terms is **478.5**.

Would you like details or have any questions?

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Here are 8 related questions you might find interesting:
1. How do you find the common difference in an AP if only the first term and the 10th term are given?
2. What is the formula for the sum of the first \( n \) terms of an AP?
3. How does changing the common difference affect the terms in an AP?
4. Can the common difference in an AP be a fraction?
5. What happens if the common difference is negative?
6. How would you calculate the 50th term of an AP?
7. What is the difference between an arithmetic progression and a geometric progression?
8. How can you determine if a given sequence is an AP?

**Tip**: When solving arithmetic progression problems, always write down the general formula for the \( n \)-th term and the sum of the first \( n \) terms. This helps in setting up equations quickly.

Learn how to solve arithmetic progression (AP) problems involving the sums of the 5th and 13th terms, the determination of the 40th term, and the sum of the 29th term. Understand the formulas for nth term and sum of first n terms of an AP. Suitable for high school students.

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