# Short Note on Basics of Arithmetic Progression

By Gaurav Mohanty|Updated : December 21st, 2021

BBA Entrance Exam focuses on the application of basic concepts. Arithmetic Progression is an important topic for BBA & IPM Entrance Exam. Therefore, we at BYJU'S Exam Prep is here to help you with the basic concepts of Arithmetic Progression. In this article, you would find certain illustrations related to the basic concepts of Arithmetic Progression.

### PROGRESSIONS

It is not necessary that the terms of a sequence always follow a certain pattern or they are described by some explicit formula for the nth term. Those sequences whose terms follow certain patterns are called progressions.

#### ARITHMETIC PROGRESSION (AP)

A sequence of numbers is said to be in Arithmetic progression if each number differs from its succeeding number by a constant value.

Common difference: Take the series: 1, 4, 7, 10, …… In this, every term increases by the same number every time. Any term deducted from the succeeding term results in the same difference of 3. (4 - 1 = 3 and 7 - 4 = 3). This is called the common difference and is denoted by 'd'.

The AP could also have a negative common difference: 6, 3, 0, -3…..each term is reduced by 3 successively. Here the common difference is -3.

The AP can be denoted by a, a + d, a + 2d, a + 3d ……where ‘a’ is the first term and ‘d’ is a common difference.

If the number of terms in the sequence = n and its first term is a, then the nth term of the AP is given by: a + (n - 1) d where d is a common difference.

The sum of n terms of an AP is given by the equation

We can also say that the Sum = (n/2)(a + L)  where L is the last term of the series.

Illustration 1: Find the 20th term of series 4, 7, 10…… and also the sum of the 20 terms.

Solution:

Here a = 4, d = 3, n = 20.

20th term is given by a + (n – 1)d ⇒ 4 + 19 × 3 = 61.

Sum of 20 terms = 20/2 × [8 + 19 × 3] = 650.

Illustration 2: The sum of certain number of terms of an A.P. is 57 and the first and last terms are 17 and 2 respectively. Find the number of terms and the common difference of the series.

Solution:

Here S = 57, a = 17, l = 2.

Using the formula S = n/2 (a + l) we get 57 = n/2 (17 + 2) ⇒ 114 = 19n, or n = 6.

Since the 6th term is 2, we can use the formula for nth term: a + (n – 1)d ⇒ 2 = 17 + 5 × d; On solving we get 5d = - 15 and d = -3.

Illustration 3: The ratio of the sum of n terms of two A.P.’s is (7n + 1) : (4n + 27). Find the ratio of their mth terms.

Solution:

Let a1, a2 be the first terms and d1, d2 the common differences of the two given A.P.’s.

Then sums of their n terms are given by

To find the ratio of the mth terms of the two given A.P.’s.

We replace n by (2m – 1) in (iii)

Hence the ratio of the mth terms of the two A.P.’s is (14m – 6) : (8m + 23).

Illustration 4: The ratio of the sums of m and n terms of an A.P. is m2 : n2. Find the ratio of the mth and nth terms.

Solution: Let a be the first term and d the common difference of the given A.P. Then, the sums of m and n terms are given by  and  respectively. Then,

⇒ {2a+(m-1)d} n={2a+(n-1)d}m

⇒ 2a(n-m) = d {(n-1)m - (m-1)n}

⇒ 2a(n-m) = d(n-m) ⇒ d=2a

#### HOW TO CHOOSE TERMS IN AN A.P.:

Sometimes we require certain number of terms in A.P. The following ways of selecting terms are generally very convenient.

In case of an odd number of terms, the middle term is ‘a’ and the common difference is ‘d’ while in even number of terms the middle terms are a – d, a + d and the common differences is 2d.

#### IMPORTANT PROPERTIES OF AP

1. If a constant is added to or subtracted from each term of an A.P., then the resulting sequence is also an A.P. with the same common difference.
2. In each term of a given A.P. is multiplied or divided by a non-zero constant k, then the resulting sequence is also an A.P. with common difference kd or d/k, where d is the common difference of the given A.P.
3. In a finite A.P. the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of first and last term i.e., ak + an (k – 1) = a1 + an for all k = 1, 2, 3, …, n – 1.
4. Three numbers a, b, c are in A.P. if 2b = a + c.

Illustration 5: If log10 2, log10 (2x – 1) and log10 (2x + 3) are in A.P., then find the value of x.

Solution:

log10 2, log10 (2x – 1) and log10 (2x + 3) are in A.P.

⇒ 2 log10 (2x – 1) = log10 2 + log10 (2x + 3)

⇒ log10 (2x – 1)2 = log10 2 . (2x + 3)

⇒ (2x – 1)2 = 2 . (2x + 3)

⇒ (y – 1)2 = 2 (y + 3), where y = 2x

⇒ y2 – 4y – 5 = 0

⇒ (y – 5) (y + 1) = 0

⇒ y = 5 or y = – 1

⇒ Hence 2x = 5; or x = log2 5

1. A sequence is an A.P. if the sum of its first n terms is of the form An2 = Bn, where A, B are constants independent of n. In such a case the common difference is 2A.

#### INSERTION OF ARITHMETIC MEANS

If 15, 11, 7, 3, –1 are in A.P., it follows that 11, 7, 3 are three arithmetic means between 15 and – 1.

If a, b, c are in A.P., then ‘b’ is the arithmetic mean of ‘a’ and ‘c’.

Illustration 6: Insert three arithmetic means between 3 and 19.

Solution:

Difference = 19 – 3 = 16

Since three arithmetic means are to be inserted, we see that common difference d = 16/(3+1) = 4.

Hence, the required A.M.’s are 7, 11, 15.

==========================================