To solve this problem, let's use the formula for the $n$th term and the sum of the first $n$ terms of an Arithmetic Progression (AP).

Step 1: Define the terms of the Arithmetic Progression

Let:

• $a$ be the first term of the AP.
• $d$ be the common difference.

The $n$th term of the AP is given by: $T_n = a + (n-1)d$

Step 2: Use the given information

We are given the sums of the 5th and 13th terms:

• $T_5 + T_{13} = 21$
• $T_5 + T_{13} = 37$

Using the formula for the $n$th term: $T_5 = a + 4d$ $T_{13} = a + 12d$

Now, substitute into the given equations:

1. $T_5 + T_{13} = 21$: $(a + 4d) + (a + 12d) = 21$ $2a + 16d = 21$ $a + 8d = \frac{21}{2} = 10.5 \quad \text{(Equation 1)}$

2. $T_5 + T_{13} = 37$: $(a + 4d) + (a + 12d) = 37$ $2a + 16d = 37$ $a + 8d = \frac{37}{2} = 18.5 \quad \text{(Equation 2)}$

Step 3: Solve the simultaneous equations

We have two equations: $a + 8d = 10.5 \quad \text{(Equation 1)}$ $a + 8d = 18.5 \quad \text{(Equation 2)}$

It seems like there was a mistake in the problem setup, as both equations are derived from the same relationship. Let's clarify by revisiting the conditions.

If these equations were correct, there would be a contradiction because they can't hold true simultaneously. Since the problem might have meant different equations or conditions, let's assume the error was in transcribing the problem. Otherwise, the problem doesn't have a consistent solution with the information given. Could you clarify or confirm the details?

Step 4: Find the 40th term and the sum of the 29th term

If we resolve the equations correctly, we'd find $a$ and $d$, then:

1. The 40th term ($T_{40}$): $T_{40} = a + 39d$

2. The sum of the first 29 terms: $S_{29} = \frac{29}{2} \times (2a + 28d)$

If you clarify the given information or if you realize the original equations are conflicting, I can proceed with accurate calculations.

Would you like to clarify the problem details or proceed with the setup?