In this module, we will learn about the �n^th term of an AP (Arithmetic Progression)�.

What is Arithmetic Progression?

Arithmetic Progression is a list of numbers in which each term except the first term is the result of adding the same number, called the common difference, to the preceding term.

Let a₁, a₂, a₃ ... be an AP whose first term a₁ is a and the common difference is d.

Then,

The 2^nd term a₂ = a + d = a + (2 � 1) d

The 3^rd term a₃ = a₂ + d = (a + d) + d = a + 2d = a + (3 � 1) d

The 4^th term a₄ = a₃ + d = (a + 2d) + d = a + 3d = a + (4 � 1) d

. . . . . . . .

. . . . . . . .

Looking at the pattern, we can say that the

n^th term a_n = a + (n � 1) d

So, the n^th term a_n of the AP with first term a and common difference d is given by

a_n = a + (n � 1) d

a_n is also called the general term of the AP. If there are m terms in the AP, then a_m represents the last term which is sometimes also denoted by l.

The nth term of an AP(Arithmetic Progression)

Let us take some examples for better understanding of nth term of an AP.

Example: Find the 8^th term of the AP: 3, 7, 11 ...

Solution:

Here, a = 3, d = 7 � 3 = 4 and n = 8

We have,

a_n = a + (n � 1) d

So,

a₈ = 3 + (8 � 1) � 4

    = 3 + 7 � 4

    = 3 + 28 = 31 Therefore, the 8^th term of the given AP is 31.

Example: Find the AP whose 4^th term is 26 and the 10^th term is 56.

Solution:

We have,

a_n = a + (n � 1) d

So,

a₄ = a + (4 � 1) d = a + 3d = 26 � (1)

And a₁₀ = a + (10 � 1) d = a + 9d = 56 � (2)

Solving the pair of linear equations (1) and (2),

a + 3d = 26 � (1)

a + 9d = 56 � (2)

We get a = 11, d = 5

Hence, the required AP is:

11, 11 + 5, 11 + 5 + 5 ...

= 11, 16, 21 �

Example: 200 cubes are stacked in the following manner: 20 cubes in the bottom row, 19 in the next row, 18 in the row next to it and so on. How many cubes are in the 10^th row?

Solution:

Number of cubes in first, second, third and next rows are 20, 19, 18, 17 �.

Since the cubes are decreased by 1 in each above row.

So the number of cubes in each row make an AP:

20, 19, 18, 17 �

Here a = 20, d = 19 � 20 = -1

We want to find the number of cubes in 10^th row, so here n = 10

We have,

a_n = a  + (n � 1) d

So, a₁₀ = a + (10 � 1) � (-1)

           = 20 + 9 � (-1)

           = 20 � 9 = 11

Therefore, 11 cubes will be required in the 10^th row.

Let's discuss �Sum of first n terms of an AP�.

Suppose we want to find the sum of the first 100 natural numbers.

We can write the sum of first 100 natural numbers as:

S = 1 + 2 + 3 + � + 98 + 99 + 100 � (i)

And then, reversed the numbers to write:

S = 100 + 99 + 98 + . . . + 3 + 2 + 1 � (ii)

Adding these two, we get

2S = (100 + 1) + (99 + 2) + (98 + 3) ... + (3 + 98) + (2 + 99) + (1 + 100)

���� = 101 + 101 + . . . + 101 + 101 (100 times) So, S = (100X100)/2= 5050

We will now use the same technique to find the sum of the first n terms of an AP:

a, a + d, a + 2d, ...

The n^th term of this AP is a + (n � 1) d.

Let S denote the sum of the first n terms of the AP.

We have,

S = a + (a + d) + (a + 2d) + ... + [a + (n � 1) d]                 � (i)

Rewriting the terms in reverse order, we have

S = [a + (n � 1) d ] + [a + (n � 2) d] + ... + (a + d ) + a     � (ii)

On adding (i) and (ii), term-wise, we get

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

2S = n [2a + (n � 1) d]

S = n/2[2a + (n � 1) d]

So, the sum of the first n terms of an AP is given by

S = n/2[2a + (n � 1) d]

We can also write as

S = n/2[a + a + (n � 1) d]
S = n/2[a + an] � (iii)

If there are only n terms in an AP, then a_n = l, the last term.

From (iii),

S = n/2[a + l] � (iv)

This formula is useful when the first and the last terms of an AP are given and the common difference is not given.

Example: Find the sum of the first 30 terms of the AP: 7, 12,17 ...

Solution: Here, a = 7, d = 12 � 7 = 5, n = 30

We know that

S = n/2[2a + (n � 1) d]

Therefore,

S = 30/2[2 � 7 + (30 � 1) � 5]

   = 15 [14 + 29 � 5]

   = 15 [14 + 145] = 15 � 159 = 2385

So, the sum of the first 30 terms of the AP is 2385.

Example: Find the sum of the first 20 terms of the arithmetic series if a₁ = 5 and a₂₀ = 62.

Solution: Here, a₁ = 5 and a₂₀ = 62 and n = 20

If we have given the first term and the last term, then

S = n/2[a + a_n]

Therefore,

S = 20/2[5 + 20]

   = 10 � 25 = 250

So, the sum of the first 20 terms of the AP is 250.

Example: How many terms of the AP: 24, 21, 18 ... must be taken so that their sum is 78?

Solution: Here, a = 24, d = 21 � 24 = �3, S_n = 78. We need to find n.

We know that

S_n = n/2[2a + (n � 1) d]

Therefore,

78 = n/2[2 � 24 + (n � 1) � (-3)]
78 � 2 = n [48 � 3n + 3]
156 = 51n � 3n²
3n² � 51n + 156 = 0
n² � 17n + 52 = 0 (Dividing throughout 3)
n² � 13n � 4n + 52 = 0
n (n � 13) � 4 (n � 13) = 0
(n � 4) (n � 13) = 0
n � 4 = 0 or n � 13 = 0
n = 4 or n = 13

Both values of n are admissible. So, the number of terms is either 4 or 13.