Arithmetic Progression- Definition and formulas

Arithmetic Progression:

Consider the series 1, 5, 9, 13, 17 or a series containing the terms 10, 16, 22, 28. Are you able to observe the pattern emerging in the above series? In both the examples given above, the difference between any two adjacent terms is the same. Such a series in which the common difference between any two adjacent terms is the same is known as an arithmetic progression.

Let us learn various formulas and their derivations now.

Formulas in Arithmetic Progression

General Expression of arithmetic progression

Let us start with the general form of an arithmetic progression. An AP can be expressed as a, a+d, a+2d, a+3d….. a+(n-1)d

Where,

a= first term

d= common difference

n= number of terms in the arithmetic progression.

Nth term of an AP Series

In an arithmetic progression where the first term and the common difference is known, the nth term of the series can be calculated using the formula T_n = a+(n-1)d

For example, we can calculate the 12th term of an AP having the first term a = 5 and the common difference d= 7 by substituting these values in the above formula and we get 82 as the 12^th term of the series.

Similarly, if we are to calculate the nth term from the end of the AP, then we use the formula
T_n = l-(n-1)d

Where

l =  the last term

n = number of terms

d = the common difference

Number of terms in an AP

When the first and last terms of an AP and the common difference is known, then we can calculate the number of terms in the arithmetic progression using the formula

n= ((l-a))/d+1

Where,

l= the last term of the AP

a= the first term of the AP

d= the common difference

n= number of terms in the AP

Sum of a finite series

When we know the first term and the common difference in the AP, the formula for calculating the sum of an arithmetic progression is S= (n[2a+(n-1)d])/2

When we know the first and the last term, the sum of n terms in an AP is S= (n(a+l))/2

Where a is the first term and l is the last term.

Derivation

Let us say there are n terms in an arithmetic progression having the common difference d. The sum of such AP series would be S=a + (a+d)+ (a+2d)+ ….. [a+(n-1)d]            (1)

Writing the same in the reverse order would be 

S= [a+(n-1)d] + [a+(n-2)d] + [a+(n-3)d]…. (a+d) + a                                           (2)

Adding (1) and (2) we get,

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

I.e. 2S =n[2a+(n-1)d]

I.e.

The sum of first n terms S=(n[2a+(n-1)d])/2..

Formula summary

Solved Examples

1. In an AP 5, 13, 21, 29, ….. 261, find the following:
   1. Number of terms in the AP
   2. 8^th term from the beginning of the AP
   3. 8^th term from the end of the AP

Solution:

We can calculate the number of terms in an AP using the formula n= ((l-a))/d+1

In the given question, the first term a = 5, last term l = 261, and the common difference d = 8.

Substituting these values, we get n= ((261-5))/8+1

Thus, the number of terms in the AP is 33.

We know that we can calculate the nth term of an AP using the formula T_n=a+(n-1)d.

Substituting the values given in the above formula,

T₈=5+(8-1)8

T₈=61

Using the formula Tn = l-(n-1)d we can calculate the 8th term from the end of the AP.

Substituting the figures from the question, we get Tn = 261-(8-1)8

Thus, Tn = 205

2. If the 8th and 10th terms of an AP are 57 and 71 respectively, find the 15th term of the AP 

Solution:

We know that the nth term of an AP is Tn =a+(n-1)d

So, we get T8 =57=a+(8-1)d

 57=a+7d (1)

Similarly, T10 =71=a+(10-1)d

71=a+9d (2)

Subtracting (1) from (2), we get,

14 = 2d.

d= 7

Substituting the value of d = 7 in (2), we get

71=a+9×7

a=8

Now, T12=8+(12-1)7

T12 =85

3. Find the sum of the first 20 terms of the AP 2, 7, 12, 17….

Solution:

From the given question, we know that a=2 and d= ( 7-2)=5.

We know that the sum of first n numbers in an AP is S= (n[2a+(n-1)d])/2

Substituting the values in the formula, we get

S₂₀= (20[2✕2+(20-1)✕5])/2

S₂₀= (20[4+95])/2

S₂₀=