Geometric progressions · 3 marks

Check whether the following sequences are geometric progressions and find their nth terms.

Answer: All three are GPs. (i) r = 5, t_n = 2 × 5ⁿ⁻¹ (ii) r = 23, t_n = 4 × (23)ⁿ⁻¹ (iii) r = −12, t_n = 3 × (−12)ⁿ⁻¹

Step-by-step solution

Idea: Find the ratio of each term to the one before. If it is constant, the sequence is a GP and t_n = arⁿ⁻¹.

GP; t_n = 2 × 5ⁿ⁻¹

83 ÷ 4 = 23; 169 ÷ 83 = 169 × 38 = 23; 3227 ÷ 169 = 3227 × 916 = 23. Constant, so a GP with a = 4, r = 23.½ mark
t_n = 4 × (23)ⁿ⁻¹.½ mark

GP; t_n = 4 × (23)ⁿ⁻¹

−32 ÷ 3 = −12; 34 ÷ −32 = 34 × 2−3 = −12; −38 ÷ 34 = −12. Constant, so a GP with a = 3, r = −12.½ mark
t_n = 3 × (−12)ⁿ⁻¹.½ mark

GP; t_n = 3 × (−12)ⁿ⁻¹

(i) GP, tₙ = 2 × 5ⁿ⁻¹ (ii) GP, tₙ = 4 × (2/3)ⁿ⁻¹ (iii) GP, tₙ = 3 × (−1/2)ⁿ⁻¹

Check: (i) n = 4: 2 × 125 = 250 ✓ (ii) n = 4: 4 × 827 = 3227 ✓ (iii) n = 4: 3 × (−18) = −38 ✓.

(i)

102 = 5010 = 25050 = 5 (constant) ∴ GP; a = 2, r = 5

∴ t_n = 2 × 5ⁿ⁻¹

(ii)

83 ÷ 4 = 169 ÷ 83 = 3227 ÷ 169 = 23 (constant) ∴ GP; a = 4, r = 23

∴ t_n = 4 × (23)ⁿ⁻¹

(iii)

−32 ÷ 3 = 34 ÷ −32 = −38 ÷ 34 = −12 (constant) ∴ GP; a = 3, r = −12

∴ t_n = 3 × (−12)ⁿ⁻¹

In (iii), dropping the minus sign and giving r = 12. The terms alternate in sign, so r is negative.
In (ii), taking r = 32 by dividing the earlier term by the later one.
Writing (iii) as 3 × −12ⁿ⁻¹ without brackets; the whole of −12 is raised to the power.

MCQ (1 mark)

The nth term of the GP 5, −15, 45, −135, … is

Show answer

(B) 5 × (−3)ⁿ⁻¹
a = 5, r = −155 = −3.

Check whether 81, 27, 9, 3, … is a GP and find its nth term.

Show answer

2781 = 927 = 39 = 13: GP. nth term = 81 × (13)ⁿ⁻¹.