MP Board · Class 10 · Mathematics · Arithmetic ProgressionsDerive the formula for the sum of the first $n$ terms of an Arithmetic Progression whose first term is '$a$' and common difference is '$d$'. Also, state the alternative formula when the last term is given.

Introduction to Sum of an AP\nLet an Arithmetic Progression have its first term as '$a$', common difference as '$d$', and number of terms as '$n$'. Let $S_n$ denote the sum of the first $n$ terms of this AP.

Representation of the Sum\nThe $n$ terms of the AP can be written as:

$$S_n = a + (a + d) + (a + 2d) + \dots + [a + (n - 2)d] + [a + (n - 1)d]$$\nHere, the last term (let's call it $l$ or $a_n$) is given by $a + (n - 1)d$.

Derivation Process

1. Write the sum in the forward order: $$S_n = a + (a + d) + (a + 2d) + \dots + [a + (n - 2)d] + [a + (n - 1)d] \quad \text{--- (Equation 1)}$$

2. Write the sum in reverse order: $$S_n = [a + (n - 1)d] + [a + (n - 2)d] + \dots + (a + d) + a \quad \text{--- (Equation 2)}$$

3. Add Equation 1 and Equation 2 term by term: $$2S_n = [a + a + (n - 1)d] + [(a + d) + a + (n - 2)d] + \dots + [a + (n - 1)d + a]$$

4. Simplifying each bracket gives: $$2S_n = [2a + (n - 1)d] + [2a + (n - 1)d] + \dots + [2a + (n - 1)d] \quad \text{(Total } n \text{ terms)}$$

5. Since the term $[2a + (n - 1)d]$ is added $n$ times: $$2S_n = n \times [2a + (n - 1)d]$$

6. Dividing by 2, we get the standard formula: $$S_n = \frac{n}{2} [2a + (n - 1)d]$$

Alternative Formula\nThe formula can be split as:

$$S_n = \frac{n}{2} [a + a + (n - 1)d]$$\nSince $a + (n - 1)d = l$ (where $l$ is the last term), the alternative formula is: $$S_n = \frac{n}{2} (a + l)$$

Conclusion\nThese formulas provide a quick and efficient way to calculate the sum of any given number of terms in an arithmetic sequence.