Algebra: Arithmetic Progression

A sequence of numbers is said to be an arithmetic progression if the difference of any two successive terms in the sequence is always a constant. This constant difference is called “common difference” of the arithmetic progression. Usually, the first term and the common difference of the arithmetic sequence are denoted by a and d respectively.

i.e. if a₁, a₂, …, a_n is the arithmetic sequence then

a=a1 and

d= a₂-a₁ = a₃-a₂ = ….= Any term – previous term.

So, we can re-write the above arithmetic progression as,

a, a+d, a+2d,…, where

the n^th term can be written as T_n = a+ (n-1)d

Example:

2,5,8,… is an arithmetic progression because 5-2=3, 8-5=3….

Here, a=2, d=3.

Here nth term = 2+(n-1)3 = 2+ 3n – 3 = 3n -1.

Sum of n elements of arithmetic progression:

It is easy to find the sum of all elements in an arithmetic sequence using normal addition if the number of elements is less. But some times this is very difficult to find the sum if there are more number of elements in the sequence. For this, let us derive a formula for finding the sum of n elements in an arithmetic sequence.

Let us consider the sequence a, a+d, a+2d,….

Let S_n be the sum of n elements of this sequence.

Then,

S_n = a+(a+d)+ (a+2d)+…+ [a+(n-1)d] … (1)

By re-writing Sn from backwards,

S_n = [a+(n-1)d] + [a+(n-2)d] + … + (a+d) + a … (2)

Adding (1) and (2),

2S_n = [2a+(n-1)d] + [2a+(n-1)d] + …. + [2a+(n-1)d]

2S_n = n[2a+(n-1)d]

S_n =n/2 *  [2a+(n-1)d]

Example:

Find the sum of first 10 elements of the sequence 2,5,8,….

Here a=2, d=3, n=10.

So, S₁₀ = 10/2 * [2*2 + (10-1) 3]

= 5 *[4 + 27] = 5 * 31 = 151.

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