Arithmetic Progressions Class 10: NCERT Notes

Arithmetic progression(AP) also called an arithmetic sequence is a sequence of numbers such that the difference from any succeeding term to its preceding term remains constant throughout the sequence

A progression is a sequence or series of numbers in which they are arranged in a particular order such that the relation between the consecutive terms of a series or sequence is always constant. In a progression, it is possible to obtain the n_th term of the series.

In mathematics, there are 3 types of progressions:

1. Arithmetic Progression (AP)
2. Geometric Progression (GP)
3. Harmonic Progression (HP)

let’s learn about AP in this article.

Arithmetic Progression (AP) also known as Arithmetic Sequence is a sequence or series of numbers such that the common difference between two consecutive numbers in the series is constant.

For example:

• Series 1: 1,3,5,7,9,11….

In this series, the common difference between any two consecutive numbers is always 2.

• Series 2: 28,25,22,19,16,13….

In this series, the common difference between any two consecutive numbers is strictly -3.

Terminology and Representation in Arithmetic Progressions

• Common difference, d = a₂ – a₁ = a₃ – a₂ = ……. = a_n – a_{n – 1}
• a_n = n_th term of Arithmetic Progression
• S_n = Sum of first n elements in the series

If a is taken as the first term and d is taken as the common difference, then the N^th term of the AP will be given by the formula: 

By computing the n terms of an AP with the upgiven formula, the general form of the AP is as follows:

Example: Find the 35^th term of the series 5,11,17,23…..

Solution:

In the given series,

a = 5, d = a₂ – a₁ = 11 – 5 = 6, n = 35

We have to find out the 35^th term, hence, apply the formulae,

a_n = a + (n – 1)d

a_n = 5 + (35 – 1) x 6

a_n = 5 + 34 x 6

a_n = 209

Hence 209 is the 35_th term.

The formula for the arithmetic progression sum is,

S_n = (n/2)[2a + (n – 1) × d]

S_n = (n/2)[a + l]

where,

• a is the First Term of Series
• l is the Last Term of Series
• n is the Number of Terms in Series

Derivation of Formula

Let ‘l’ denote the n_th term of the series and S_n be the sum of first n terms of AP a, (a+d), (a+2d), …., a+(n-1)d then,

S_n = a₁ + a₂ + a₃ + ….a_n-1 + a_n

S_n = a + (a + d) + (a + 2d) + …….. + (l – 2d) + (l – d) + l                      …(1)

Writing the series in reverse order, we get,

S_n = l + (l – d) + (l – 2d) + …….. + (a + 2d) + (a + d) + a                      …(2)

Adding equation (1) and (2),

2S_n = (a + l) + (a + l) + (a + l) + …….. + (a + l) + (a + l) + (a + l)

2S_n = n(a + l)

S_n = (n/2)(a + l)             …(3)

Hence, the formulae for finding the sum of a series is,

S_n = (n/2)(a + l) 

Replacing the last term l by the n^th term in equation 3 we get,

n^th term = a + (n – 1)d

S_n = (n/2)(a + a + (n – 1)d)

S_n = (n/2)(2a + (n – 1) x d)

Note: Consecutive terms in an Arithmetic Progression can also be represented as,

…….., a-3d , a-2d, a-d, a, a+d, a+2d, a+3d, ……..

Related Article:

• Summation Formula
• Sum of Natural Numbers
• Arithmetic Progression and Geometric Progression

Problem 1. Find the sum of the first 35 terms of series 5,11,17,23…..

Solution:

In the given series,

a = 5, d = a₂ – a₁ = 11 – 5 = 6, n = 35

S_n = (n/2)(2a + (n – 1) x d)

S_n = (35/2)(2 x 5 + (35 – 1) x 6)

S_n = (35/2)(10 + 34 x 6)

S_n = (35/2)(10 + 204)

S_n = 35 x 214/2

S_n = 3745

Problem 2. Find the sum of the series when the first term of the series is 5 and the last of the series is 209 and the number of terms in the series is 35.

Solution:

In the given series,

a = 5, l = 209, n = 35

S_n = (n/2)(a + l)

S_n = (35/2)(5 + 209)

S_n = 35 x 214/2

S_n = 3745

Problem 3. 21 Rupees is divided among three brothers where the three parts of money are in AP and the sum of their squares is 155. Find the largest amount.

Solution:

Let the money is deivided in three parts as (a-d), a, (a+d)

Given,

(a – d) + a + (a + d) = 21

Therefore, 

3a = 21

a = 7

Again, (a – d)² + a² + (a + d)² = 155

a² + d² – 2ad + a² + a² + d² + 2ad = 155

3a² + 2d² = 155

Putting the value of ‘a’ we get,

3(7)² + 2d² = 155

2d² = 155 – 147

d² = 4

d = ±2

Three parts of distributed money are:

a + d = 7 + 2 = 9

a = 7

a – d = 7 – 2 = 5

Hence, the largest part is Rupees 9

Q1. Sum of the first 12 terms of an arithmetic progression is equal to the sum of the next 12 terms. Prove that the 13th term is zero.

Q2. In an arithmetic progression, if the sum of the first p terms is equal to the sum of the first q terms, prove that the (p+q)^th term is zero.

Q3. The 3rd term of an arithmetic progression is 4, and the 8th term is -11. Find the 20th term.

Q4. If the sum of the first n terms of an arithmetic progression is S_n = n/2.(3n+1), find the 15th term.

Q5. Sum of the first n terms of an arithmetic progression is S_n = 5n² – n. Find the common difference and the 10th term.

What is the mean of Arithmetic Progression?

Arithmetic progression or AP in mathematics is defined as a sequence in which the difference between any two consecutive number of the sequence is always constant. Example, 2, 4, 6, 8, … is an AP as the difference between any two consecutive terms in the sequence is always 2.

What is nth term of an AP?

N^th(T_n) term of the AP is calculated using the formula:

T_n = a + (n – 1)d

What is the formula for sum of n term of an AP?

Sum of ‘n’ terms of an AP (S_n) is calculated using the formula:

S_n = n/2{2a + (n – 1)d}

Is a constant sequence always an AP?

Yes, contact sequences such as, 1, 1, 1,…, are always AP as their common difference is always constant(zero).