How To Calculate and Solve Arithmetic Progression In Sequences and Series

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 T1, T2, T3, T4, T5, … is a linear sequence, the common difference is obtained as:

T2 – T1 = T3 – T2 = T4 – T3 = T5 – T4
Therefore, d = Tn – T(n – 1)
where
Tn = nth term of the sequence and T(n – 1) is the (n – 1)th</sup< term.

Let’s take a quick example:
Lets find the common difference in the following arithmetic progression:
a) 0,5,10,15,20
b) 314, 512, 734, 10, 1214

Solution

a) To find the common difference we use the formula, Tn – T(n – 1)
d = T2 – T1
T2 = 5 and T1 = 0
d = 5 – 0
d = 5
Therefore, the common difference for the first series (a) is 5.

b) Applying the same principle,
d = T2 – T1
T2 = 51 d = 512 – 314
d = 214

The nth term of an arithmetic progression is denoted as Tn or Un,
if T1</sub = a
T2 = T1 + d = a + d
T3 = T2 + d = a + d + d = a + 2d

It implies that
Tn = a + (n – 1)d
This formula above is the general formula used in solving arithmetic progression

Let’s take some examples:
a) Find the nth term in 34, 40, 46
b) The second, third and fourth terms in an arithmetic progression are x – 2, 5 and x + 2 respectively 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 nth term of the following arithmetic progression 32, 29, 26, …, -118
e) The 8th term of an arithmetic progression is 18 and the 12th term is 26. Find the first term, common difference and the 20th term.
f) The first term of an arithmetic progression is 10, the ratio of the 7th term to the 9th term is 3:5. Calculate the common difference of the progression.
g) The numbers 11, x, y, 2112 from an arithmetic progression. Find the values of x and y.

Solution
a) We have the formula as Tn = a + (n – 1)d
a = 34
d = T2 – T1 = 40 – 34 = 6
Therefore,
Tn = 34 + (n – 1)6
Tn = 34 + 6n – 6
Tn = 34 – 6 + 6n
Tn = 28 + 6n

b)
2nd term = x – 2
3rd term = 5
4th term = x + 2

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

d = T3 – T2 = T4 – T3
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
Tn = l (Last Term) = 31x
d (Common Difference) = 5x

Tn = a + (n – 1)d
31x = x + (n – 1)5x
31x = x + 5nx – 5x
31x = -4x + 5nx
31x + 4x = 5nx
35x = 5nx
7 = n
Therefore, the number of terms is 7.

d)
A.P. => 32, 29, 26, …, -118
Let’s first of all find the number of terms
General Formula: Tn = a + (n – 1)d
a = 32
d = T2 – T1
d = 29 – 32 = -3
l = -118

-118 = 32 + (n – 1)-3
-118 = 32 -3n + 3
-118 = 35 – 3n
3n = 35 + 118
3n = 153
n = 51
Therefore, the number of terms is 51.
Next, we find the nth term expression
Tn = 32 + (n – 1)-3
Tn = 32 + 3 – 3n
Tn = 35 – 3n

e)
T8 = 18, n = 8
T12 = 26, n = 12

First, let’s find the T8 and T12 expression
T8 = a + (8 – 1)d
T8 = a + 7d … (I)

T12 = a + (12 – 1)d
T12 = a + 11d … (II)

Next, 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
26 = a + 11(2)
26 = a + 22
a = 26 – 22
a = 4
First Term is 4

Next, we calculate the value of the 20th term
T20 = 4 + (20 – 1)2
T20 = 4 + 19(2)
T20 = 4 + 38
T20 = 42

f)
First Term (a) = 10
T7 : T9 = 3:5 = 35
Using the general formula: Tn = a + (n – 1)d

T7 = 10 + (7 – 1)d
T7 = 10 + 6d

T9 = 10 + (9 – 1)d
T9 = 10 + 8d

Since T7:T9 = 3:5
It implies that 5T7 = 3T9

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, 2112
General Formula: Tn = a + (n – 1)d
T4 = 2112, n = 4, a = 11

2112 = 11 + (4 – 1)d
2112 = 11 + 3d
432 = 11 + 3d
Multiplying throughout by 2
43 = 22 + 6d
43 – 22 = 6d
21 = 6d
d = 21/6
d = 3.5
Next, Let’s find x

T2 = x = 11 + (2 – 1)3.5
x = 11 + (1)3.5
x = 11 + 3.5
x = 14.5

Next, Let’s find y

T3 = y = 11 + (3 – 1)3.5
y = 11 + (2)3.5
y = 11 + 7
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) = T2 – T1
d (Common Difference) = 4 – 2 = 2

Tn = a + (n – 1)d
100 = 2 + (n – 1)2
100 = 2 + 2n – 2
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

Adding the two series vertically we have:

2S = 102 + 102 + … + 102 + 102 + 102
2S = 102 x 50
2S = 5100
S = 51002
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)

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
Substituting l = a + (n – 1)d into equation (V)
Sn = n(a + l)2
Sn = n(a + a + (n – 1)d)2
Sn = 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 one more example

a) The first term and last term of an arithmetic progression are 2 and 256 respectively. 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
Sn (Sum of the Series) = 1548

Sn = n(a + l)2
1548 = n(2 + 256)2
Multiplying throughout by 2
3096 = n(2 + 256)
3096 = 258n
n = 3096258
n = 12

Therefore, the number of terms of the series is 12

Next, Let’s find the common difference, d

Sn = n(2a + (n – 1)d)2
1548 = 12(2(2) + (12 – 1)d)2
Multiplying throughout by 2
3096 = 12(4 + (11)d)
3096 = 48 + 132d
3096 – 48 = 132d
3048 = 132d
d = 3048132
d = 23.0909

Therefore, the common difference is 23.0909.

Nickzom Calculator is capable of solving all the parameters of an Arithmetic Progression.

Download Our Free Android App

21 thoughts on “How To Calculate and Solve Arithmetic Progression In Sequences and Series”

  1. 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

    1. 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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