Chapter 5: Arithmetic Progressions

We have a₁ = – 1 , a₂ = – 1, a₃ = – 1 and a₄ = – 1

a₂ – a₁ = 0

a₃ – a₂ = 0

a₄ – a₃ = 0

Clearly, the difference of successive terms is same, therefore given list of numbers from an AP.

(ii) 0, 2, 0, 2,…

We have a₁ = 0, a₂ = 2, a₃ = 0 and a₄ = 2

a₂ – a₁ = 2

a₃ – a₂ = – 2

a₄ – a₃ = 2

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(iii) 1, 1, 2, 2, 3, 3…

We have a₁ = 1 , a₂ = 1, a₃ = 2 and a₄ = 2

a₂ – a₁ = 0

a₃ – a₂ = 1

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(iv) 11, 22, 33…

We have a₁ = 11, a₂ = 22 and a₃ = 33

a₂ – a₁ = 11

a₃ – a₂ = 11

Clearly, the difference of successive terms is same, therefore given list of numbers form an AP.

(v) 1/2,1/3,1/4, …

We have a₁ = ½ , a₂ = 1/3 and a₃ = ¼

a₂ – a₁ = -1/6

a₃ – a₂ = -1/12

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(vi) 2, 2², 2³, 2⁴, …

We have a₁ = 2 , a₂ = 2², a₃ = 2³ and a₄ = 2⁴

a₂ – a₁ = 2² – 2 = 4 – 2 = 2

a₃ – a₂ = 2³ – 2² = 8 – 4 = 4

Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(vii) √3, √12, √27, √48, …

We have,

a₁ = √3, a₂ = √12, a₃ = √27 and a₄ = √48

a₂ – a₁ = √12 – √3 = 2√3 – √3 = √3

a₃ – a₂ = √27 – √12 = 3√3 – 2√3 = √3

a₄ – a₃ = √48 – √27 = 4√3 – 3√3 = √3

Clearly, the difference of successive terms is same, therefore given list of numbers from an AP.

False

a₁ = -1, a₂ = -3/2, a₃ = -2 and a₄ = 5/2

a₂ – a₁ = -3/2 – (-1) = – ½

a₃ – a₂ = – 2 – (- 3/2) = – ½

a₄ – a₃ = 5/2 – (-2) = 9/2

Clearly, the difference of successive terms in not same, all though, a₂ – a₁ = a₃ – a₂ but a₄ – a₃ ≠ a₃ – a₂ therefore it does not form an AP.

True

Given

First term, a = – 3

Common difference, d = a₂ – a₁ = – 7 – (- 3) = – 4

a₃₀ – a₂₀ = a + 29d – (a + 19d)

= 10d

= – 40

It is so because difference between any two terms of an AP is proportional to common difference of that AP

Suppose there are two AP’s with first terms a and A

And their common differences are d and D respectively

Suppose n be any term

a_n = a + (n – 1)d

A_n = A + (n – 1)D

As common difference is equal for both AP’s

We have D = d

Using this we have

A_n – a_n = a + (n – – 1)d – [ A + (n – 1)D]

= a + (n – 1)d – A – (n – 1)d

= a – A

As a – A is a constant value

Therefore, difference between any corresponding terms will be equal to a – A.