RD Sharma Class 10 Solutions Chapter 9 - Exercise 9.3

While studying the arithmetic progression, we may come across the following terms. 

First Term: As given by its name, the first term in an AP is the first term in the progression. It is usually represented by a1 or simply a. For example, in sequence 2,4,6,8,10…. 2 is the first term. 

Common Difference: As mentioned above, an AP is a sequence where every term, except the first term, is obtained by adding a fixed number to its previous term. This fixed number is called the “common difference” and it is generally denoted by 'd'. If the first term is a1, then: the second term is a1+d, the third term is a1+d+d = a1+2d, and the fourth term is a1+2d+d= a1+3d and the sequence goes on. For example, in the sequence 2,4,6,8,10. . . Each term, except the first term which is 2, is obtained by adding 2 to its previous term. So, in this case, the common difference is, d=2. Generally, the common difference is the difference between every two successive terms of an Arithmetic progression. Hence, the formula to calculate the common difference of an AP is:  d=an-a{n-1}

The General Term of an Arithmetic Progression (nth Term)

The general term also called the nth term of an AP, whose first term is a and the common difference is d, is found by the formula an=a+(n-1)d. 

The Formula for Calculating Sum of Arithmetic Progression

Consider an arithmetic progression sequence whose first term is a1 or a and the common difference is d.

• The sum of the first n terms of an arithmetic progression when the nth term is not known to us is Sn = n/2[2a+(n-1) d] 

• The sum of the first n terms of an arithmetic progression when the nth term, an is known to us is Sn = n/2[a1+an] 

Derivation of the Arithmetic Progression Formula

To find the nth term of an arithmetic progression, we know that the nth term, an=a1+(n–1)d, where, a1 is the first term, a1 + d is the second term, the third term is a1 + 2d, and the sequence goes on. For finding the sum of the arithmetic series, say Sn, we start with the first term and successively add the common difference to the next coming terms.

Sn = a1 + (a1 + d) + (a1 + 2d) + … + [a1 + (n–1)d].

Also, we can start with the nth term and successively subtract the common difference.

Sn = an + (an – d) + (an – 2d) + … + [an – (n–1)d].

So, the sum of the arithmetic sequence could be found in any of the two ways. However, 

on adding those two equations together, we will be getting

Sn = a1 + (a1 + d) + (a1 + 2d) + … + [a1 + (n–1)d]

Sn = an + (an – d) + (an – 2d) + … + [an – (n–1)d]

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2Sn = (a1 + an) + (a1 + an) + (a1 + an) + … + [a1 + an]

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We can notice that all the terms that have d are added out. Hence,

2Sn = n (a1 + an)

Sn = [n(a1 + an)]/2

By substituting an = a1 + (n – 1)d into the last formula, we have

Sn = n/2 [a1 + a1 + (n – 1)d] ...Simplifying

Sn = n/2 [2a1 + (n – 1)d].

These two formulas help us to find the sum of an arithmetic sequence quickly.