# How To Calculate and Solve Arithmetic Progression In Sequences and Series

**Last Updated on February 5, 2024**

## Arithmetic Progression

A sequence is a set of numbers arranged in a definite pattern. Each number is called a term.

A finite sequence is one that has a last term when listed. For Example: 2,4,6,8,,…,16. An infinite sequence is one that does not have a last term when listed.

Arithmetic Progression follows the rule of linear sequence which is the sequence in which each term is obtained by adding a distant number (Positive or Negative) to the proceeding terms.

The constant number is called **common difference** and it is denoted as “d” while the **first term** is denoted as “a”.

If T_{1}, T_{2}, T_{3}, T_{4}, T_{5}, … is a linear sequence, the common difference is obtained as:

T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} = T_{5} – T_{4}

Therefore, **d** = T_{n} – T_{(n – 1)}

where

T_{n} = n^{th} term of the sequence and T_{(n – 1)} is the (n – 1)^{th</sup< term.}

In other words, here is a quick example:

Lets find the common difference in the following arithmetic progression:**a)** 0,5,10,15,20**b)** 3^{1}⁄_{4}, 5^{1}⁄_{2}, 7^{3}⁄_{4}, 10, 12^{1}⁄_{4}

**Solution**

a) To find the **common difference** we use the formula, T_{n} – T_{(n – 1)}

d = T_{2} – T_{1}

T_{2} = 5 and T_{1} = 0

d = 5 – 0

d = 5

Therefore, the common difference for the first series (a) is 5.

b) Then, by applying the same principle,

d = T_{2} – T_{1}

T_{2} = 5^{1 d = 51⁄2 – 31⁄4d = 21⁄4}

The n^{th} term of an arithmetic progression is denoted as **T _{n}** or

**U**,

_{n}if T

_{1</sub = aT2 = T1 + d = a + dT3 = T2 + d = a + d + d = a + 2d}

Therefore, this implies that

T_{n} = a + (n – 1)d

Hence, this formula above is the general formula used in solving arithmetic progression

##### Let’s take some examples:

**a)** Find the n^{th} term in 34, 40, 46**b)** The second, third and fourth terms in an arithmetic progression are x – 2, 5 and x + 2 respectively. Therefore, calculate the value of x.**c)** Find the number of terms in an arithmetic progression given that its first and last terms are x and 31x and its common difference is 5x.**d)** Find the number of terms and the expression for the n^{th} term of the following arithmetic progression 32, 29, 26, …, -118**e)** The 8^{th} term of an arithmetic progression is 18 and the 12^{th} term is 26. So, find the first term, common difference and the 20^{th} term.**f)** The first term of an arithmetic progression is 10, the ratio of the 7^{th} term to the 9^{th} term is 3:5. Then, calculate the common difference of the progression.**g)** The numbers 11, x, y, 21^{1}⁄_{2} from an arithmetic progression. Hence, find the values of x and y.

**Solution**

a) We have the formula as T_{n} = a + (n – 1)d

a = 34

d = T_{2} – T_{1} = 40 – 34 = 6

Therefore, T_{n} = 34 + (n – 1)6

T_{n} = 34 + 6n – 6

T_{n} = 34 – 6 + 6n

i.e. T_{n} = 28 + 6n

b)

2^{nd} term = x – 2

3^{rd} term = 5

4^{th} term = x + 2

##### Since there is no first term, we would use common difference to calculate the value of x

d = T_{3} – T_{2} = T_{4} – T_{3}

d = 5 – (x – 2) = (x + 2) – 5

5 – x + 2 = x + 2 – 5

5 + 2 – 2 + 5 = x + x

10 = 2x

5 = x

Therefore, the value of x is 5.

c)

a (First Term) = x

T_{n} = l (Last Term) = 31x

d (Common Difference) = 5x

T_{n} = a + (n – 1)d

31x = x + (n – 1)5x

31x = x + 5nx – 5x

i.e. 31x = -4x + 5nx

31x + 4x = 5nx

35x = 5nx

7 = n

Therefore, the number of terms is 7.

d)

A.P. => 32, 29, 26, …, -118

So, let’s first of all find the number of terms

General Formula: T_{n} = a + (n – 1)d

a = 32

d = T_{2} – T_{1}

d = 29 – 32 = -3

l = -118

-118 = 32 + (n – 1)-3

-118 = 32 -3n + 3

So, -118 = 35 – 3n

3n = 35 + 118

3n = 153

n = 51

Therefore, the number of terms is 51.

Then, we find the n^{th} term expression

T_{n} = 32 + (n – 1)-3

T_{n} = 32 + 3 – 3n

i.e. T_{n} = 35 – 3n

e)

T_{8} = 18, n = 8

T_{12} = 26, n = 12

First, let’s find the T_{8} and T_{12} expression

T_{8} = a + (8 – 1)d

T_{8} = a + 7d … (I)

T_{12} = a + (12 – 1)d

T_{12} = a + 11d … (II)

So, we obtain the values of a and d using simultaneous equation for equation (I) and (II)

**Using Elimination Method**

18 = a + 7d

-26 = a + 11d

This leaves us with this equation

-8 = -4d

d = -8 / -4 = 2

Common Difference is 2

##### Using Equation (II) to find the value of a (first term)

26 = a + 11d

That is, 26 = a + 11(2)

26 = a + 22

a = 26 – 22

a = 4

First Term is 4

Ne, we calculate the value of the 20^{th} term, which will be T_{20} = 4 + (20 – 1)2

T_{20} = 4 + 19(2)

T_{20} = 4 + 38

That is, T_{20} = 42

f)

First Term (a) = 10

T_{7} : T_{9} = 3:5 = ^{3}⁄_{5}

So, by using the general formula: T_{n} = a + (n – 1)d

T_{7} = 10 + (7 – 1)d

T_{7} = 10 + 6d

Then,

T_{9} = 10 + (9 – 1)d

T_{9} = 10 + 8d

Since T_{7}:T_{9} = 3:5

It implies that 5T_{7} = 3T_{9}

5(10 + 6d) = 3(10 + 8d)

50 + 30d = 30 + 24d

30d – 24d = 30 – 50

6d = -20

d = -3.3333

Therefore, the common difference is -3.3333

g)

A.P. => 11, x, y, 21^{1}⁄_{2}**General Formula:** T_{n} = a + (n – 1)d

T_{4} = 21^{1}⁄_{2}, n = 4, a = 11

21^{1}⁄_{2} = 11 + (4 – 1)d

21^{1}⁄_{2} = 11 + 3d^{43}⁄_{2} = 11 + 3d

Multiplying throughout by 2

43 = 22 + 6d

43 – 22 = 6d

21 = 6d

d = 21/6

d = 3.5

Then, Let’s find x

T_{2} = x = 11 + (2 – 1)3.5

x = 11 + (1)3.5

x = 11 + 3.5

That is, x = 14.5

So, let’s find y

T_{3} = y = 11 + (3 – 1)3.5

y = 11 + (2)3.5

y = 11 + 7

Therefore, y = 18

**Series**

A series is obtained when the terms of a sequence are added.**For Example:**

a) 9 + 12 + 15 + 18 + … + 51

b) 4 + 8 + 12 + .. + 48

c) 1 + 2 + 3 + 4 + …

d) 5 + 10 + 15 + 20 + …

Examples a) and b) are called **finite series** because they have a definite number of terms and it is always possible to find the **sum of a finite series**.

Examples c) and d) are called **infinite series**. The sum of infinite series is often impossible to find.

**Let’s solve something**

Find the sum of this **A.P.**

2 + 4 + 6 + … + 98 + 100**Solution**

a (First Term) = 2, l (Last Term) = 100

d (Common Difference) = T_{2} – T_{1}

d (Common Difference) = 4 – 2 = 2

T_{n} = a + (n – 1)d

100 = 2 + (n – 1)2

100 = 2 + 2n – 2

This means 100 = 2n

n = 100/2

n = 50

Therefore, the number of terms in the arithmetic progression is **50**

**Take a closer look at this**

S (Sum) = 2 + 4 + 6 + … + 98 + 100

Reversing the Series

S (Sum) = 100 + 98 + … + 6 + 4 + 2

**So, by adding the two series vertically we have:**

2S = 102 + 102 + … + 102 + 102 + 102

2S = 102 x 50

Hence, 2S = 5100

S = ^{5100}⁄_{2}

S = 2550

Therefore, the sum of the arithmetic series is **2550**

The following expression represents a general arithmetic series, where the terms are added.

S = a + (a + d) + (a + 2d) + … + l … (III)

S = l + (l – d) + (l – 2d) + … + a … (IV)

##### So, by adding the equations (III) and (IV)

2S = (a + l) + (a + l) + … + (a + l)

2S = (a + l)n

S = ^{n(a + l)}⁄_{2} … (V)

The above formula is used when first term and last term is given

There is another formula

Since l = a + (n – 1)d

Then substituting l = a + (n – 1)d into equation (V)

S_{n} = ^{n(a + l)}⁄_{2}

S_{n} = ^{n(a + a + (n – 1)d)}⁄_{2}

Therefore, S_{n} = ^{n(2a + (n – 1)d)}⁄_{2} … (VI)

The above equation or equation (VI) is used when the first term and common difference is given.

**Let’s take another example**

**a)** The first term and last term of an arithmetic progression are 2 and 256 respectively. Hence, how many terms are there in the sequence and what is the common difference of the series if its sum is 1548?

**Solution**

a)

First, Let’s find the number of terms, n

a (First Term) = 2

l (Last Term) = 256

S_{n} (Sum of the Series) = 1548

S_{n} = ^{n(a + l)}⁄_{2}

1548 = ^{n(2 + 256)}⁄_{2}

Then, multiplying throughout by 2

3096 = n(2 + 256)

3096 = 258n

n = ^{3096}⁄_{258}

n = 12

Therefore, the number of terms of the series is **12**

So, let’s find the common difference, d

S_{n} = ^{n(2a + (n – 1)d)}⁄_{2}

1548 = ^{12(2(2) + (12 – 1)d)}⁄_{2}

So, multiplying throughout by 2

This will give you 3096 = 12(4 + (11)d)

3096 = 48 + 132d

3096 – 48 = 132d

3048 = 132d

d = ^{3048}⁄_{132}

d = 23.0909

Therefore, the common difference is **23.0909**.

Nickzom Calculator is capable of solving all the parameters of an **A**rithmetic **P**rogression.

Pls can you help me solve this arithmetic progression……the cost per meter of sinking a well increases with the depth.for the first meter it is #50.30,for the second meter it is #57.90, for the third it is #65.50 etc.find the maximum depth possible if the cost is not to exceed #20,000

First, you find the common difference using Nickzom Calculator

You entered 65.5 as the value for the nth term.

You entered 3 as the value for the term’s position.

You entered 50.3 as the value for the first term

Steps:

d = (Un – a) / (n – 1)

d = (65.5 – 50.3) / (3 – 1)

d = (15.200000000000003) / (2)

d = 7.600000000000001

Now, you find the term’s position at 20,000 still using Nickzom Calculator

You entered 20000 as the value for the nth term.

You entered 50.3 as the value for the first term.

You entered 7.6 as the value for the common difference

Steps:

n = ((Un – a) / d) + 1

n = ((20000 – 50.3) / 7.6) + 1

n = ((19949.7/ 7.6) + 1)

n = 2624.9605263157896 + 1

n = 2625.9605263157896

Therefore, the maximum depth possible if not to exceed #20,000 is 2625.

Link to the app in Nickzom Calculator is https://www.nickzom.org/calculator/aripro.php

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