Arithmetic Progression: Introduction & Solved Examples | Quantitative Aptitude (Quant) - CAT PDF Download

What is Arithmetic Progression?

A progression is a special type of sequence for which it is possible to obtain a formula for the nth term.  The Arithmetic Progression is the most commonly used sequence in maths with easy to understand formulas. 

Definition 1: A mathematical sequence in which the difference between two consecutive terms is always a constant and it is abbreviated as AP.

Definition 2: An arithmetic sequence or progression is defined as a sequence of numbers in which for every pair of consecutive terms, the second number is obtained by adding a fixed number to the first one.

The fixed number that must be added to any term of an AP to get the next term is known as the common difference of the AP. Now, let us consider the sequence, 1, 4, 7, 10, 13, 16,…

It is considered as an arithmetic sequence (progression) with a common difference 3. 

Notation in Arithmetic Progression

In AP, we will come across some main terms, which are denoted as:

• First term (a)
• Common difference (d)
• nth Term (a_n)
• Sum of the first n terms (S_n)

All three terms represent the property of Arithmetic Progression. We will learn more about these three properties in the next section.

First Term of AP

The AP can also be written in terms of common differences, as follows;

a, a + d, a + 2d, a + 3d, a + 4d, ………. ,a + (n – 1) d

where  “a” is the first term of the progression. 

Common Difference in Arithmetic Progression

In this progression, for a given series, the terms used are the first term, the common difference and nth term. Suppose, a₁, a₂, a₃, ……………., a_n is an AP, then; the common difference “ d ” can be obtained as;

d = a₂ – a₁ = a₃ – a₂ = ……. = a_n – a_{n – 1}

Where “d” is a common difference. It can be positive, negative or zero.

General Form of an AP

Consider an AP to be: a₁, a₂, a₃, ……………., a_n

Arithmetic Progression Formulas

• The nth term of AP
• Sum of the first n terms

Solved Examples

Below are the problems to find the nth term and the sum of the sequence, which are solved using AP sum formulas in detail. Go through them once and solve the practice problems to excel in your skills.

Example 1: Find the value of n, if a = 10, d = 5, a_n = 95.

Solution: Given, a = 10, d = 5, a_n = 95
From the formula of general term, we have:
a_n = a + (n − 1) × d
95 = 10 + (n − 1) × 5
(n − 1) × 5 = 95 – 10 = 85
(n − 1) = 85/ 5
(n − 1) = 17
n = 17 + 1
n = 18

Example 2: Find the 20th term for the given AP:3, 5, 7, 9, ……

Solution: Given, 
3, 5, 7, 9, ……
a = 3, d = 5 – 3 = 2, n = 20
a_n = a + (n − 1) × d
a₂₀ = 3 + (20 − 1) × 2
a₂₀ = 3 + 38
⇒a₂₀ = 41

Example 3: Find the sum of the first 30 multiples of 4.

Solution:The first 30 multiples of 4 are: 4, 8, 12, ….., 120
Here, a = 4, n = 30, d = 4
We know,
S30 = n/2 [2a + (n − 1) × d]
S30 = 30/2[2 (4) + (30 − 1) × 4]
S30 = 15[8 + 116]
S30 = 1860

Example 4.:Find the series whose n^th term is . Is it an A. P. series? If yes, find 101^st term. 

Solution: Putting 1, 2, 3, 4…. We get T1, T2, T3, T4…………..

As the common differences are equal
∴The series is an A.P.

Example 5: A student purchases a pen for Rs. 100. At the end of 8 years, it is valued at Rs. 20. Assuming
 that the yearly depreciation is constant. Find the annual depreciation.

Solution: Original cost of pen = Rs. 100
Let D be the annual depreciation.
∴ Price after one year = 100 - D = T₁ = a (say)
∴ Price after eight years = T₈ = a + 7 (- D) = a - 7D
= 100 - D - 7D = 100 - 8D
By the given condition 100 - 8D = 20
8D = 80

∴D = 10.
Hence annual depreciation = Rs. 10.