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|Arithmetic Progressions Questions and Solutions|
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Arithmetic Progression: A sequence of numbers in which the successive terms increase or decrease by a constant number is called an “Arithmetic Progression (AP)”
Common Difference: The constant difference between two consecutive terms of an AP is called “Common Difference”.
General Form of an AP: If the first term is ‘a’ and the common difference is ‘d’, then a, a + d, a + 2d, a + 3d,... represent the arithmetic progression for different values of ‘a’ and ‘d’. This is called “General Form of an AP”
the nth term of an A.P.: The nth term of an AP is given by:
an = a + (n - 1)d.
Sum of first n -terms of an AP: If a, a + d, a + 2d, ... is an AP and nth term an = a+(n-1) d. Then, the sum of n-terms is: Sn = n/2 [2a + (n - 1)d] or Sn = n/2 (a + 1)
where a = First term of an AP
l = Last term of an AP = a + (n - 1) d
Arithmetic Progression (AP) is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value. For example, the series of natural numbers: 1, 2, 3, 4, 5, 6,… is an AP, which has a common difference between two successive terms (say 1 and 2) equal to 1 (2 -1). Even in the case of odd numbers and even numbers, we can see the common difference between two successive terms will be equal to 2.
If we observe in our regular lives, we come across Arithmetic progression quite often. For example, Roll numbers of students in a class, days in a week or months in a year. This pattern of series and sequences has been generalized in Maths as progressions.
In mathematics, there are three different types of progressions. They are:
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. Let’s have a look at its three different types of definitions.
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.
Definition 3: 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,… is considered as an arithmetic sequence with common difference 3.
In AP, we will come across three main terms, which are denoted as:
All three terms represent the property of Arithmetic Progression. We will learn more about these three properties in the next section.
In this progression, for a given series, the terms used are the first term, the common difference between the two terms and nth term. Suppose, a1, a2, a3, ……………., an is an AP, then; the common difference “ d ” can be obtained as;
d = a2 – a1 = a3 – a2 = ……. = an – an – 1
Where “d” is a common difference. It can be positive, negative or zero.
First Term of AP
The AP can also be written in terms of common difference, as follows;
a, a + d, a + 2d, a + 3d, a + 4d, ………. ,a + (n – 1) d
where “a” is the first term of the progression.
General Form of an A. P
Consider an AP to be: a1, a2, a3, ……………., an
Position of Terms
Representation of Terms
Values of Term
a = a + (1-1) d
a + d = a + (2-1) d
a + 2d = a + (3-1) d
a + 3d = a + (4-1) d
a + (n-1)d
Let us learn here both the formulas with examples.
nth Term of an AP
an = a + (n − 1) × d
a = First term
d = Common difference
n = number of terms
an = nth term
Example: Find the nth term of AP: 1, 2, 3, 4, 5…., an, if the number of terms are 15. Solution: Given, AP: 1, 2, 3, 4, 5…., an
n = 15
By the formula we know, an = a + (n - 1)d
First-term, a = 1 Common difference, d = 2 - 1 = 1
Therefore, an = 1 + (15 - 1)1 = 1 + 14 = 15
Note: The finite portion of an AP is known as finite AP and therefore the sum of finite AP is known as arithmetic series. The behaviour of the sequence depends on the value of a common difference.
For any progression, the sum of n terms can be easily calculated. For an AP, the sum of the first n terms can be calculated if the first term and the total terms are known. The formula for the arithmetic progression sum is explained below:
Consider an AP consisting “n” terms.
S = n/2[2a + (n − 1) × d]
This is the AP sum formula to find the sum of n terms in series.
Proof: Consider an AP consisting “n” terms having the sequence a, a + d, a + 2d, ………….,a + (n – 1) × d
Sum of first n terms = a + (a + d) + (a + 2d) + ………. + [a + (n – 1) × d] ——————-(i)
Writing the terms in reverse order, we have:
S = [a + (n – 1) × d] + [a + (n – 2) × d] + [a + (n – 3) × d] + ……. (a) ———–(ii)
Adding both the equations term wise, we have:
2S = [2a + (n – 1) × d] + [2a + (n – 1) × d] + [2a + (n – 1) × d] + …………. + [2a + (n – 1) ×d] (n-terms)
2S = n × [2a + (n – 1) × d]
S = n/2[2a + (n − 1) × d]
Example: Let us take the example of adding natural numbers up to 15 numbers.
AP = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
Given, a = 1, d = 2 - 1 = 1 and an = 15
Now, by the formula we know;
S = n/2[2a + (n − 1) × d] = 15/2[2.1 + (15 - 1).1]
S = 15/2[2 + 14] = 15/2  = 15 x 8
S = 120
Hence, the sum of the first 15 natural numbers is 120.
Formula to find the sum of AP when first and last terms are given as follows:
S = n/2 (first term + last term)
The list of formulas is given in a tabular form used in AP. These formulas are useful to solve problems based on the series and sequence concept.
|General Form of AP||a, a + d, a + 2d, a + 3d, . . .|
|The nth term of AP||an = a + (n – 1) × d|
|Sum of n terms in AP||S = n/2[2a + (n − 1) × d]|
|Sum of all terms in a finite AP with the last term as ‘l’||n/2(a + l)|
Below are the problems to find the nth terms and sum of the sequence are solved using AP sum formulas in detail. Go through them once and solve the practice problems to excel your skills.
Example 1: Find the value of n. If a = 10, d = 5, an = 95.
Solution: Given, a = 10, d = 5, an = 95
From the formula of general term, we have:
an = 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, ……
3, 5, 7, 9, ……
a = 3, d = 5 – 3 = 2, n = 20
an = 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: Given, a = 4, n = 30, d = 4
S = n/2 [2a + (n − 1) × d]
S = 30/2[2 (4) + (30 − 1) × 4]
S = 15[8 + 116]
S = 1860