Which of the following list of numbers does form an AP? If they form a

1.	Which of the following list of numbers does form an AP? If they form an AP, write the next two terms :(i) 4, 10, 16, 22, . . . (ii) 1, – 1, – 3, – 5, . . . (iii) – 2, 2, – 2, 2, – 2, . . . (iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . .
Answer» (i) We have a₂ – a₁ = 10 – 4 = 6 a₃ – a₂ = 16 – 10 = 6 a₄ – a₃ = 22 – 16 = 6 i.e., a_{k + 1} – a_k is the same every time. So, the given list of numbers forms an AP with the common difference d = 6. The next two terms are: 22 + 6 = 28 and 28 + 6 = 34. (ii) a₂ – a₁ = – 1 – 1 = – 2 a₃-a₂=-3-(-1)=-3+1=-2 a₄-a₃= -5-(-3)=-5+3=-2 i.e., a_k_{+ 1} – a_k is the same every time. So, the given list of numbers forms an AP with the common difference d = – 2. The next two terms are: – 5 + (– 2 ) = – 7 and – 7 + (– 2 ) = – 9 (iii) a₂ – a₁ = 2 – (– 2) = 2 + 2 = 4 a₃ – a₂ = – 2 – 2 = – 4 As a₂-a₁ a₃-a₂, the given list of numbers does not form an AP. (iv) a₂– a₁ = 1 – 1 = 0 a₃ – a₂ = 1 – 1 = 0 a₄-a₃=2-1=1 3 = 2 – 1 = 1 Here, a₂ – a₁ = a₃– a₂ ≠a₄ – a₃. So, the given list of numbers does not form an AP.

Which of the following list of numbers does form an AP? If they form an AP, write the next two terms :(i) 4, 10, 16, 22, . . . (ii) 1, – 1, – 3, – 5, . . . (iii) – 2, 2, – 2, 2, – 2, . . . (iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . .

Answer»

(i) We have a₂ – a₁ = 10 – 4 = 6

a₃ – a₂ = 16 – 10 = 6

a₄ – a₃ = 22 – 16 = 6

i.e., a_{k + 1} – a_k is the same every time.

So, the given list of numbers forms an AP with the common difference d = 6.

The next two terms are: 22 + 6 = 28 and 28 + 6 = 34.

(ii) a₂ – a₁ = – 1 – 1 = – 2

a₃-a₂=-3-(-1)=-3+1=-2

a₄-a₃= -5-(-3)=-5+3=-2

i.e., a_k_{+ 1} – a_k is the same every time.

So, the given list of numbers forms an AP with the common difference d = – 2.

The next two terms are:

– 5 + (– 2 ) = – 7 and – 7 + (– 2 ) = – 9

(iii) a₂ – a₁ = 2 – (– 2) = 2 + 2 = 4

a₃ – a₂ = – 2 – 2 = – 4

As a₂-a₁ a₃-a₂, the given list of numbers does not form an AP.

(iv) a₂– a₁ = 1 – 1 = 0

a₃ – a₂ = 1 – 1 = 0

a₄-a₃=2-1=1

3 = 2 – 1 = 1

Here, a₂ – a₁ = a₃– a₂ ≠a₄ – a₃.

So, the given list of numbers does not form an AP.