Question

Show that a₁, a₂, a₃, … a₁₁. … form an AP where a₁₁ is defined as below: 

(i) a = 3 + 4n 

(ii) a = 9 – 5n

Answer 1

(i) a_n= 3 + 4n 

a₁ = 3 + 4(1) = 3 + 4 

∴ a_n = 7 

∴ a = 7 

a₁ = 3 + 4n 

a₂ = 3 + 4 × 2 = 3 + 8 

∴ a₂ = 11 

a_n = 3 + 4n 

a₃ = 3 + 4 × 3 

= 3 + 12 

a₃ = 15 

∴ a₁, a₂, a₃, …………. 

7, 11, 15, ………. 

d = a₂ – a₁ = 11 – 7 = 4 

d = a₃ – a₂= 15 – 11 = 4 

Here, the value of ‘d’ is constant. 

∴ a_n = 3 + 4n forms an Arithmetic Progression. 

(ii) a_n = 9 – 5n 

a₁ = 9 – 5 × 1 = 9 – 5 

∴ a₁ = 4 

a₁ = 9 – 5n 

a₁ = 9 – 5 × 2 

= 9 – 10 

∴ a₂ = -1 

a_n= 9 – 5n 

a₃ = 9 – 5 × 3 

= 9 – 15 

∴ a₃ = -6 

a₁, a₂, a₃, …………… a_n 

4, -1, -6, ………… 

d = a₂ – a₁ = -1 – 4 = -5 

d = a₃ – a₂ = -6 – (-1) = -5 

Here, the value of ‘d’ is constant. 

∴ a_n = 9 – 5n form an Arithmetic Progression.