Class 10 Maths Arithmetic Progressions Notes

5.1 Introduction to Arithmetic Progressions

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. This difference is called the common difference, denoted by ddd.

For example:

  • 2,5,8,11,14,…2, 5, 8, 11, 14, \dots2,5,8,11,14,… is an arithmetic progression with the common difference d=3d = 3d=3.
  • 20,17,14,11,8,…20, 17, 14, 11, 8, \dots20,17,14,11,8,… is also an AP, but with a common difference d=βˆ’3d = -3d=βˆ’3.

The general form of an arithmetic progression is:a,a+d,a+2d,a+3d,…a, a + d, a + 2d, a + 3d, \dotsa,a+d,a+2d,a+3d,…

Where:

  • aaa is the first term,
  • ddd is the common difference,
  • The nnn-th term is represented as an=a+(nβˆ’1)β‹…da_n = a + (n – 1) \cdot dan​=a+(nβˆ’1)β‹…d.

Arithmetic progressions are widely used in real-life situations, such as in financial calculations, construction, and even in computer algorithms.


5.2 Arithmetic Progressions: Key Concepts

An Arithmetic Progression (AP) has the following key properties:

  • The first term of an AP is denoted by aaa.
  • The common difference ddd is the difference between any two consecutive terms.
  • The general form of an arithmetic progression is:

a,a+d,a+2d,a+3d,…a, a + d, a + 2d, a + 3d, \dotsa,a+d,a+2d,a+3d,…

Where:

  • aaa is the first term,
  • ddd is the common difference.

The nnn-th term of an AP is given by the formula:an=a+(nβˆ’1)β‹…da_n = a + (n – 1) \cdot dan​=a+(nβˆ’1)β‹…d

Where:

  • ana_nan​ is the nnn-th term,
  • nnn is the position of the term in the sequence.

Example:

If the first term a=2a = 2a=2 and the common difference d=3d = 3d=3, the first few terms of the AP are:2,5,8,11,14,…2, 5, 8, 11, 14, \dots2,5,8,11,14,…

For this sequence:

  • The 1st term (a1a_1a1​) is 2,
  • The 2nd term (a2a_2a2​) is 5,
  • The 3rd term (a3a_3a3​) is 8,
    and so on.

5.3 nth Term of an AP

The nnn-th term of an arithmetic progression can be found using the formula:an=a+(nβˆ’1)β‹…da_n = a + (n – 1) \cdot dan​=a+(nβˆ’1)β‹…d

Where:

  • ana_nan​ is the nnn-th term,
  • aaa is the first term,
  • ddd is the common difference,
  • nnn is the position of the term in the progression.

Example:

Find the 10th term of the arithmetic progression 3,6,9,12,15,…3, 6, 9, 12, 15, \dots3,6,9,12,15,….

  • Here, a=3a = 3a=3 and d=3d = 3d=3.
  • Using the formula for the nnn-th term:

a10=3+(10βˆ’1)β‹…3=3+27=30a_{10} = 3 + (10 – 1) \cdot 3 = 3 + 27 = 30a10​=3+(10βˆ’1)β‹…3=3+27=30

So, the 10th term of this AP is 30.

Another Example:

For the AP 5,8,11,14,…5, 8, 11, 14, \dots5,8,11,14,…, find the 7th term.

  • a=5a = 5a=5 and d=3d = 3d=3.
  • Using the formula:

a7=5+(7βˆ’1)β‹…3=5+18=23a_7 = 5 + (7 – 1) \cdot 3 = 5 + 18 = 23a7​=5+(7βˆ’1)β‹…3=5+18=23

So, the 7th term of this AP is 23.


5.4 Sum of First n Terms of an AP

The sum of the first nnn terms of an arithmetic progression is given by the formula:Sn=n2β‹…[2a+(nβˆ’1)β‹…d]S_n = \frac{n}{2} \cdot \left[ 2a + (n – 1) \cdot d \right]Sn​=2n​⋅[2a+(nβˆ’1)β‹…d]

Alternatively, the sum can also be expressed as:Sn=n2β‹…(a+l)S_n = \frac{n}{2} \cdot (a + l)Sn​=2n​⋅(a+l)

Where:

  • SnS_nSn​ is the sum of the first nnn terms,
  • aaa is the first term,
  • lll is the last term of the progression,
  • ddd is the common difference.

Example:

Find the sum of the first 10 terms of the AP 2,5,8,11,…2, 5, 8, 11, \dots2,5,8,11,….

  • Here, a=2a = 2a=2, d=3d = 3d=3, and n=10n = 10n=10.

Using the formula for the sum of the first nnn terms:S10=102β‹…[2β‹…2+(10βˆ’1)β‹…3]S_{10} = \frac{10}{2} \cdot \left[ 2 \cdot 2 + (10 – 1) \cdot 3 \right]S10​=210​⋅[2β‹…2+(10βˆ’1)β‹…3] S10=5β‹…[4+27]=5β‹…31=155S_{10} = 5 \cdot \left[ 4 + 27 \right] = 5 \cdot 31 = 155S10​=5β‹…[4+27]=5β‹…31=155

So, the sum of the first 10 terms of the AP is 155.

Another Example:

Find the sum of the first 15 terms of the AP 7,10,13,16,…7, 10, 13, 16, \dots7,10,13,16,….

  • a=7a = 7a=7, d=3d = 3d=3, and n=15n = 15n=15.

First, find the 15th term using the nnn-th term formula:a15=7+(15βˆ’1)β‹…3=7+42=49a_{15} = 7 + (15 – 1) \cdot 3 = 7 + 42 = 49a15​=7+(15βˆ’1)β‹…3=7+42=49

Now, using the sum formula:S15=152β‹…(7+49)=152β‹…56=15β‹…28=420S_{15} = \frac{15}{2} \cdot (7 + 49) = \frac{15}{2} \cdot 56 = 15 \cdot 28 = 420S15​=215​⋅(7+49)=215​⋅56=15β‹…28=420

So, the sum of the first 15 terms of the AP is 420.


5.5 Summary

In this chapter, we have explored key concepts related to Arithmetic Progressions (AP):

  • An Arithmetic Progression is a sequence of numbers with a constant difference between consecutive terms.
  • The nnn-th term of an AP can be calculated using the formula an=a+(nβˆ’1)β‹…da_n = a + (n – 1) \cdot dan​=a+(nβˆ’1)β‹…d, where aaa is the first term, ddd is the common difference, and nnn is the position of the term.
  • The sum of the first nnn terms of an AP is given by the formula:

Sn=n2β‹…[2a+(nβˆ’1)β‹…d]S_n = \frac{n}{2} \cdot \left[ 2a + (n – 1) \cdot d \right]Sn​=2n​⋅[2a+(nβˆ’1)β‹…d]

  • Arithmetic Progressions are widely used in real-life scenarios, including calculations related to finance, construction, and more.

MCQs Based on the “Arithmetic Progressions” Chapter:

1. What is the common difference in the AP: 5,9,13,17,…5, 9, 13, 17, \dots5,9,13,17,…?

a) 2
b) 3
c) 4
d) 5

Answer: b) 4

2. In the AP 3,7,11,15,…3, 7, 11, 15, \dots3,7,11,15,…, what is the 6th term?

a) 15
b) 17
c) 19
d) 21

Answer: c) 19

3. The sum of the first 10 terms of the AP 1,4,7,10,…1, 4, 7, 10, \dots1,4,7,10,… is:

a) 150
b) 160
c) 170
d) 180

Answer: b) 160

4. The sum of the first 20 terms of the AP 2,5,8,11,…2, 5, 8, 11, \dots2,5,8,11,… is:

a) 400
b) 420
c) 450
d) 500

**Answer