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Let's go through the questions one by one and solve them.

### Q1:
Given the \(n\)-th term of an arithmetic progression (A.P.) as:
\[
A_n = (n+1)^2 - (n-1)^2
\]
Find the common difference of the A.P.

#### Solution:
Using the identity for difference of squares:
\[
A_n = (n+1)^2 - (n-1)^2 = [(n+1) - (n-1)] \cdot [(n+1) + (n-1)] = 2 \cdot (2n) = 4n
\]
Thus, the \(n\)-th term is:
\[
A_n = 4n
\]
This is a linear equation, and in an A.P., the common difference \(d\) is the difference between successive terms. So:
\[
d = A_{n+1} - A_n = 4(n+1) - 4n = 4
\]
**Answer**: The common difference \(d = 4\).

---

### Q2:
The first term of an A.P. is 5, and the sum of 15 terms is 390. Find the middle term.

#### Solution:
The sum of the first \(n\) terms of an A.P. is given by:
\[
S_n = \frac{n}{2} \cdot [2a + (n-1)d]
\]
Here, \(S_{15} = 390\), \(a = 5\), and \(n = 15\). Substituting these values:
\[
390 = \frac{15}{2} \cdot [2(5) + (15-1)d]
\]
\[
390 = \frac{15}{2} \cdot [10 + 14d]
\]
\[
390 = \frac{15}{2} \cdot (10 + 14d)
\]
\[
780 = 15 \cdot (10 + 14d)
\]
\[
52 = 10 + 14d
\]
\[
42 = 14d
\]
\[
d = 3
\]
Now, the middle term of an A.P. with an odd number of terms is the 8th term (since 15 is odd, \( \frac{15+1}{2} = 8 \)).

The 8th term is given by:
\[
A_8 = a + (8-1)d = 5 + 7 \cdot 3 = 5 + 21 = 26
\]
**Answer**: The middle term is 26.

---

### Q3:
The sum of 3 successive terms in an A.P. is 51, and the product of the first and the last term is 273. Find the terms.

#### Solution:
Let the three terms be \(a - d\), \(a\), and \(a + d\). 

- Sum of the terms:
\[
(a - d) + a + (a + d) = 3a = 51 \quad \Rightarrow \quad a = \frac{51}{3} = 17
\]
- Product of the first and last term:
\[
(a - d)(a + d) = a^2 - d^2 = 273
\]
Substitute \(a = 17\):
\[
17^2 - d^2 = 273
\]
\[
289 - d^2 = 273
\]
\[
d^2 = 16 \quad \Rightarrow \quad d = 4
\]
Thus, the terms are:
\[
a - d = 17 - 4 = 13, \quad a = 17, \quad a + d = 17 + 4 = 21
\]
**Answer**: The terms are 13, 17, and 21.

---

### Q4:
Find the number of integers between 1 and 100 divisible by 3 or 5.

#### Solution:
First, find the numbers divisible by 3, by 5, and by both (i.e., divisible by 15, which is the least common multiple of 3 and 5).

- Numbers divisible by 3: \( \left\lfloor \frac{100}{3} \right\rfloor = 33 \)
- Numbers divisible by 5: \( \left\lfloor \frac{100}{5} \right\rfloor = 20 \)
- Numbers divisible by 15: \( \left\lfloor \frac{100}{15} \right\rfloor = 6 \)

Now, use the principle of inclusion-exclusion to find the total number of integers divisible by 3 or 5:
\[
\text{Total} = 33 + 20 - 6 = 47
\]
**Answer**: There are 47 integers between 1 and 100 that are divisible by 3 or 5.

---

Let me know if you would like more details on any of these problems or their solutions. Here are some related questions for practice:

1. What is the formula for the sum of the first \(n\) terms of an A.P.?
2. How do you find the middle term in an A.P. with an even number of terms?
3. What is the general form for the \(n\)-th term of an A.P.?
4. Can you derive the inclusion-exclusion principle used in counting?
5. How would you solve for the terms if given the sum and product of 4 successive terms in an A.P.?

**Tip**: Always verify A.P. problems using the common difference and formula for the \(n\)-th term for consistency.

Solve all the arithmetic progression questions as shown in the image.

Learn how to solve arithmetic progression problems including finding the common difference, middle term, successive terms, and counting numbers divisible by 3 or 5 between 1 and 100. This solution covers key concepts like the sum of an AP series, inclusion-exclusion principle, and nth term calculation.

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