nth term of an Arithmetic Progression

n^th term of an Arithmetic Progression:

An arithmetic progression is a series, where the difference between any two consecutive numbers is the same. Consider a series containing the terms 5, 7, 9, 11, and 13. This is an arithmetic progression with a common difference of 2 and the first term is 5. The general form of an arithmetic progression is a, a+d, a+2d, and so on. Let us learn to calculate the n^th term of an arithmetic progression here.

n^th term of an Arithmetic Progression Formula:

Consider the series  5, 7, 9, 11 and 13. In this series,

T₁ = 5 = a

T₂ = 7 = a+d

T₃ = 9 = a+2d

T₄ = 11 = a+3d

We can notice a pattern emerging here. The pattern being, if we need to calculate the 3^rd term, we need to add the first term with (3-1) times common difference. Similarly, if we need to calculate the 4^th term, we need to add the first term with (4-1) times the common difference. Thus, to calculate the n^th term in an arithmetic progression, we need to add the first term with (n-1) times the common difference.

Hence, the formula to calculate the n^th term of an AP is T_n= a+(n-1)d

Now let us say we are provided with only the last term of an AP series and we are asked to calculate the n^th term from the end of the AP. To derive the formula for that, let us consider the series 5, 7, 9, 11, and 13 again, only this time, we will consider the series from the end of the series. So, T₁ will be the last term, T₂ is the second term from the end, T₃ is the third term from the end, and so on. Now we have,

T₁ = 13 = l

T₂ = 11 = l-d

T₃ = 9 = l-2d

T₄ = 7 = l-3d

We can see a pattern emerging here which is, to calculate the n^th term from the end of an AP, we have to deduct (n-1) times the common difference from the last term.

So, the formula to calculate the n^th term from the end of an AP is  T_n= l-(n-1)d

Solved Illustrations:

1. If T₅ = 70 and T₇ = 100, then find T₉ of the AP series

 Solution:

We know that T_n= a+(n-1)d

So, T₅=70= a+(5-1)d                                                                     (1)

T₇=100= a+(7-1)d                                                                         (2)

Subtracting (2) from (1), we get

30= 2d

d=15

Substituting the value of d in (1), we get

T₅=70= a+(5-1)15

a=10

Now that we know the first term and the common difference of the AP series, we can calculate the 9^th term of the AP series as below.

T₉= 10+(9-1)15

T₉= 130

2. If the n^th term of an AP is (3n + 1) then find the sum of the first four terms of the series.

Solution:

The n^th term of an AP is given as (3n + 1). Substituting 1, 2, 3, and 4 in place of n, we can get the first four terms of the AP as below

T₁ = (3(1)+1) = 4

T₂ = (3(2)+1)= 7

T₃ = (3(3)+1)= 10

T₄ = (3(4)+1) = 13

We know the first four terms of the AP series. The sum of these first four terms is 34.

3. If the last term of an AP is 86 and the common difference is 7, find the 3^rd term from the last.

Solution:

We know that the formula to calculate the n^th term from the end of an AP series is

T_n= l-(n-1)d.

Substituting the given values, we get

T₃= 86-(3-1)7

T₃= 72

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