Let's solve this step-by-step.

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

Step 1: Express the 5th and 7th Terms in Terms of $a$ and $d$

The formula for the $n$-th term of an AP is: $T_n = a + (n-1)d$

So:

• The 5th term, $T_5 = a + 4d$
• The 7th term, $T_7 = a + 6d$

According to the problem: $T_5 + T_7 = 52$ Substitute the values: $(a + 4d) + (a + 6d) = 52$ $2a + 10d = 52$ Divide by 2: $a + 5d = 26 \quad \text{(Equation 1)}$

Step 2: Express the 10th Term in Terms of $a$ and $d$

The 10th term is given as: $T_{10} = a + 9d = 46$ So: $a + 9d = 46 \quad \text{(Equation 2)}$

Step 3: Solve the Equations to Find $a$ and $d$

Now we have two equations:

1. $a + 5d = 26$
2. $a + 9d = 46$

Subtract Equation 1 from Equation 2: $(a + 9d) - (a + 5d) = 46 - 26$ $4d = 20$ $d = 5$

Substitute $d = 5$ back into Equation 1: $a + 5 \cdot 5 = 26$ $a + 25 = 26$ $a = 1$

Step 4: Write the AP and the General $n$-th Term

Now that we know $a = 1$ and $d = 5$, the AP is: $1, 6, 11, 16, 21, \ldots$

The $n$-th term of this AP is: $T_n = a + (n-1)d = 1 + (n-1) \cdot 5 = 1 + 5n - 5 = 5n - 4$

Final Answer

The AP is: $1, 6, 11, 16, 21, \ldots$

The $n$-th term is: $T_n = 5n - 4$

Would you like more details or have any questions?

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Here are some related questions for practice:

1. Find the sum of the first 20 terms of this AP.
2. Determine the 15th term of this AP.
3. If the sum of the first $n$ terms of an AP is $290$, find $n$.
4. Find the difference between the 10th and 5th terms of the AP.
5. Determine the sum of the 12th and 16th terms of this AP.

Tip: In any AP, the difference between consecutive terms is always constant, making it easy to identify the common difference by simple subtraction.