NCERT Solutions for Class 9th Maths Chapter 8 In-text Questions

Q1.

Exercise: Check whether the following sequences are geometric progressions and find their nth terms. (i) 2, 10, 50, 250, … (ii) 4, 8/3, 16/9, 32/27, … (iii) 3, –3/2, 3/4, –3/8, …

Answer

For a GP the test is on the ratios of consecutive terms, not the differences. Every consecutive pair must give the same r.

(i) 2, 10, 50, 250, …

10 ÷ 2 = 5, 50 ÷ 10 = 5, 250 ÷ 50 = 5
Constant, so it is a GP with a = 2, r = 5
t_n = 2 × 5^n–1 (t₁ = 2, t₂ = 10, t₃ = 50, t₄ = 250 ✓)

(ii) 4, 8/3, 16/9, 32/27, …

(8/3) ÷ 4 = 8/12 = 2/3
(16/9) ÷ (8/3) = (16/9) × (3/8) = 2/3
(32/27) ÷ (16/9) = (32/27) × (9/16) = 2/3
Constant, so it is a GP with a = 4, r = 2/3
t_n = 4 × (2/3)^n–1

(iii) 3, –3/2, 3/4, –3/8, …

(–3/2) ÷ 3 = –1/2
(3/4) ÷ (–3/2) = (3/4) × (–2/3) = –1/2
(–3/8) ÷ (3/4) = (–3/8) × (4/3) = –1/2
Constant, so it is a GP with a = 3, r = –1/2
t_n = 3 × (–1/2)^n–1

Why it happens: the common ratio may be greater than 1, between 0 and 1, or negative, and each choice shows on the sequence. In (i), r = 5 makes the terms race away. In (ii), 0 < r < 1 makes them shrink towards 0 while staying positive. In (iii), a negative r makes the sign flip at every step, because (–1/2)^n–1 is positive for odd n and negative for even n — the terms shrink in size while alternating in sign.

Check it yourself: in (iii), put n = 4: 3 × (–1/2)³ = 3 × (–1/8) = –3/8. ✓

Q2.

Exercise: Can you find a recursive rule for the formula tn = 3 × 10ⁿ⁻¹ that generates the geometric progression 3, 30, 300, 3000, … ?

Answer

Each term is ten times the one before, so:

t₁ = 3, t_n = 10 × t_n–1 for n ≥ 2

Running it: t₂ = 10 × 3 = 30, t₃ = 10 × 30 = 300, t₄ = 10 × 300 = 3000. ✓

Why it happens: the two rules can be shown to agree, not merely observed to. Divide the explicit rule at n by its value at n – 1:
t_n ÷ t_n–1 = (3 × 10^n–1) ÷ (3 × 10^n–2) = 10^{n–1–(n–2)} = 10.
So t_n = 10 t_n–1 for every n ≥ 2, and since t₁ = 3 × 10⁰ = 3 the two rules start at the same place as well. The same computation on t_n = ar^n–1 gives the general fact t_n = r · t_n–1.

Tip: in a GP the common ratio is a multiplier, so the recursive step is "×r", never "+r".

NCERT Solutions for Class 9th Maths Chapter 8 In-text Questions — Geometric Progressions