Question

Write the sequence with nth term:

(i) a_n = 3 + 4n

(ii) a_n = 5 + 2n

(iii) a_n = 6 - n

(iv) a_n = 9 - 5n

Show that all of the above sequences form A.P.

Answer 1

(i) Put n = 1

A₁ = 3 + 4(1) = 7

Put n = 2

A₂ = 3 + 4(2) = 11

Put n = 3

A₃ = 3 + 4(3) = 15

Common difference,

d₁ = a₂ – a₁ = 11 – 7 = 4

Common difference,

d₂ = a₃ – a₂ = 15 – 11 = 4

Since,

d₁ = d₂

Therefore, it’s an A.P. with sequence 7, 11, 15,… 

(ii) Put n = 1

A₁ = 5 + 2(1) = 7

Put n = 2

A₂ = 5 + 2(2) = 9

Put n = 3

A₃ = 5 + 2(3) = 11

Common difference,

d₁ = a₂ – a₁= 9 – 7 = 2

Common difference,

d₂ = a₃ – a₂= 11 – 9 = 2

Since,

d₁ = d₂

Therefore, it’s an A.P. with sequence 7, 9, 11,… 

(iii) Put n = 1

A₁ = 6 – 1 = 5

Put n = 2

A₂ = 6 – 2 = 4

Put n = 3

A₃ = 6 – 3 = 3

Common difference,

d₁ = a₂ – a₁= 4 – 5 = -1

Common difference,

d₂ = a₃ - a₂= 3 – 4 = -1

Since,

d₁ = d₂

Therefore, it’s an A.P. with sequence 5, 4 , 3,… 

(iv) Put n = 1

A₁ = 9 – 5(1) = 4

Put n = 2

A₂ = 9 – 5(2) = -1

Put n = 3

A₃ = 9 – 5(3) = -6

Common difference,

d₁ = a₂ – a₁ = -1 – 4 = -5

Common difference,

d₂ = a₃ – a₂ = - 6 – (-1) = - 5

Since,

d₁ = d₂

Therefore, it’s an A.P. with sequence 4, -1, -6,…