The fifth chapter of CBSE Class 10th Mathematics is ‘Arithmetic Progression’.
Students will come to know the difference between a sequence, series and progression.
A sequence is a finite or infinite list of numbers following a specific pattern. For example, 1, 2, 3, 4, 5,… is the sequence, an infinite sequence of natural numbers.
A series is the sum of the elements in the corresponding sequence. For example, 1+2+3+4+5….is the series of natural numbers. Each number in a sequence or a series is called a term.
A progression is a sequence in which the general term can be can be expressed using a mathematical formula.
An arithmetic progression (AP) is a progression in which the difference between two consecutive terms is constant. Example: 2, 5, 8, 11, 14…. is an arithmetic progression.
The difference between two consecutive terms in an AP, (which is constant) is the “common difference“(d) of an A.P. In the progression: 2, 5, 8, 11, 14 …the common difference is 3.
As it is the difference between any two consecutive terms, for any A.P, if the common difference is:
Students will come to know the difference between a sequence, series and progression.
A sequence is a finite or infinite list of numbers following a specific pattern. For example, 1, 2, 3, 4, 5,… is the sequence, an infinite sequence of natural numbers.
A series is the sum of the elements in the corresponding sequence. For example, 1+2+3+4+5….is the series of natural numbers. Each number in a sequence or a series is called a term.
A progression is a sequence in which the general term can be can be expressed using a mathematical formula.
An arithmetic progression (AP) is a progression in which the difference between two consecutive terms is constant. Example: 2, 5, 8, 11, 14…. is an arithmetic progression.
The difference between two consecutive terms in an AP, (which is constant) is the “common difference“(d) of an A.P. In the progression: 2, 5, 8, 11, 14 …the common difference is 3.
As it is the difference between any two consecutive terms, for any A.P, if the common difference is:
- positive, the AP is increasing.
- zero, the AP is constant.
- negative, the A.P is decreasing.
Some important topics to study in this chapter are as follows:
- nth term of an AP
- Sum of First n terms of an AP
Students can view and download the chapter from the link given below.
Click here to get the complete chapter
NCERT Solutions for Chapter 5: Arithmetic Progression
Also Check
Chapter 1: Real Numbers
Chapter 2: Polynomials
Chapter 3: Pair of Linear Equations in Two Variables
Chapter 4: Quadratic Equations
Chapter 6: Triangles
Chapter 7: Coordinate Geometry
Chapter 8: Introduction to Trigonometry
Chapter 9: Some Applications of Trigonometry
Chapter 10: Circle
Chapter 11: Constructions
Chapter 12: Areas Related to Circles
Chapter 13: Surface Areas and Volume
Chapter 14: Statistics
Chapter 15: Probability
Check out Frequently Asked Questions (FAQs) for Chapter 5: Arithmetic Progression
If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are
Let three numbers be a – d, a, a + d
∴ a – d +a + a + d = 9
⇒ 3a = 9
⇒ a = 3
Also (a – d) . a . (a + d) = 24
⇒ (3 -d) .3(3 + d) = 24
⇒ 9 – d² = 8
⇒ d² = 9 – 8 = 1
∴ d = ± 1
Hence numbers are 2, 3, 4 or 4, 3, 2
∴ a – d +a + a + d = 9
⇒ 3a = 9
⇒ a = 3
Also (a – d) . a . (a + d) = 24
⇒ (3 -d) .3(3 + d) = 24
⇒ 9 – d² = 8
⇒ d² = 9 – 8 = 1
∴ d = ± 1
Hence numbers are 2, 3, 4 or 4, 3, 2
The 10th term from the end of the A.P. -5, -10, -15,…, -1000 is
Here l = -1000, d = -10 – (-5) = -10 + 5 = – 5
∴ 10th term from the end = l – (n – 1 )d = -1000 – (10 – 1) (-5) = -1000 + 45 = -955
∴ 10th term from the end = l – (n – 1 )d = -1000 – (10 – 1) (-5) = -1000 + 45 = -955
Find the sum of 12 terms of an A.P. whose nth term is given by an = 3n + 4
Here an = 3n + 4
∴ a1 = 7, a2 – 10, a3 = 13
∴ a= 7, d = 10 – 7 = 3
∴ S12 = 122[2 × 7 + (12 – 1) ×3] = 6[14 + 33] = 6 × 47 = 282
∴ a1 = 7, a2 – 10, a3 = 13
∴ a= 7, d = 10 – 7 = 3
∴ S12 = 122[2 × 7 + (12 – 1) ×3] = 6[14 + 33] = 6 × 47 = 282
The sum of first n odd natural numbers is
n²
If 2x, x + 10, 3x + 2 are in A.P., then x is equal to
Since 2x, x + 10 and 3x + 2 are in A.P.
∴ 2(x + 10) = 2x + (3x + 2)
⇒ 2x + 20 – 5x + 2
⇒ 2x – 5x = 2 – 20
⇒ 3x = 18
⇒ x = 6
∴ 2(x + 10) = 2x + (3x + 2)
⇒ 2x + 20 – 5x + 2
⇒ 2x – 5x = 2 – 20
⇒ 3x = 18
⇒ x = 6