Given the arithmetic progression (A.P.), the first term is $a$, the second term is $b$, and the last term is $c$.

To find the sum of the A.P., let's derive the expression step by step.

Step 1: General terms of an A.P.

• The first term is $a$.
• The second term is $b$. The common difference $d$ can be written as:

$d = b - a$

• The $n$-th term of an A.P. is given by the formula:

$T_n = a + (n-1)d$

Since the last term is $c$, we can write:

$c = a + (n-1)d$

Substituting $d = b - a$ into the equation:

$c = a + (n-1)(b - a)$

Simplifying this equation:

$c = a + (n-1)b - (n-1)a$ $c = a + (n-1)b - (n-1)a$ $c = (n-1)(b - a) + a$

Rearranging for $n$:

$n-1 = \frac{c - a}{b - a}$ $n = \frac{c - a}{b - a} + 1$

Step 2: Sum of the A.P.

The sum $S_n$ of the first $n$ terms of an A.P. is given by the formula:

$S_n = \frac{n}{2} \times (a + l)$

where $l$ is the last term (which is $c$ here). So, substituting in the values:

$S_n = \frac{n}{2} \times (a + c)$

Now, substitute the value of $n$ from Step 1:

$S_n = \frac{\left(\frac{c - a}{b - a} + 1\right)}{2} \times (a + c)$

Simplifying this expression:

$S_n = \frac{(c - a) + (b - a)}{2(b - a)} \times (a + c)$

$S_n = \frac{b + c - 2a}{2(b - a)} \times (a + c)$

$S_n = \frac{(b + c - 2a)(a + c)}{2(b - a)}$

Thus, the sum of the A.P. is:

$S_n = \frac{(b + c - 2a)(a + c)}{2(b - a)}$

This completes the derivation.

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Would you like a more detailed explanation or further clarification on any of the steps?

Here are 5 related questions to explore this concept further:

1. How would the sum formula change if the common difference was negative?
2. What happens to the sum if $b = a$?
3. Can you derive a general formula for the sum of the first $n$ terms of an A.P. without knowing the last term?
4. What is the significance of the common difference in determining the progression of terms?
5. How does the sum of an A.P. change when the number of terms increases?

Tip: When working with arithmetic progressions, knowing the common difference is crucial to finding terms and sums easily.