Why it happens: in a GP, tm ÷ tn = rm–n — the first term cancels out. So a question that links two terms of the same GP never needs a at all, which is why the short route works and is safer arithmetic.
Q2.
Find the 10th and nth terms of the GP: 5, 25, 125, … .
Answer
a = 5 and r = 25 ÷ 5 = 5 (also 125 ÷ 25 = 5).
tn = arn–1 = 5 × 5n–1 = 5n t10 = 510 = 97,65,625
Check the rule on the printed terms: t1 = 51 = 5, t2 = 52 = 25, t3 = 53 = 125. ✓
Why it happens: here the first term happens to equal the common ratio, so 5 × 5n–1 collapses into the single power 5n by the law am × ak = am+k. This is a special feature of this GP, not a general rule — for 3, 6, 12, … the nth term stays 3 × 2n–1 and cannot be shortened.
Check it yourself: 510 = (55)2 = 31252 = 97,65,625.
Q3.
A sequence is given by the recursive rule t1 = 2, tn+1 = 3tn – 2 for n ≥ 1. Which term of the sequence is 730?
There is also a way to see it without listing. Subtract 1 from each term: 1, 3, 9, 27, 81, 243, 729 — a GP of powers of 3.
If un = tn – 1, then un+1 = tn+1 – 1 = (3tn – 2) – 1 = 3(tn – 1) = 3un with u1 = 1, so un = 3n–1 Hence tn = 3n–1 + 1 3n–1 + 1 = 730 → 3n–1 = 729 = 36 → n – 1 = 6 → n = 7
Why it happens: the rule tn+1 = 3tn – 2 is not a GP as it stands, because of the "– 2". But shifting every term down by 1 removes the constant and leaves pure tripling — the fixed point of the rule is x = 3x – 2, that is x = 1, and measuring from that fixed point turns the sequence into a GP. Recognising this converts a listing problem into a one-line equation, which is the only practical method if the question had asked about the 30th term.
Q4.
Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the nth term.
Answer
a = 2 and r = 6 ÷ 2 = 3 (also 18 ÷ 6 = 3).
Explicit: tn = 2 × 3n–1 Recursive: t1 = 2, tn = 3 × tn–1 for n ≥ 2
Now solve tn = 4374:
2 × 3n–1 = 4374 3n–1 = 2187 37 = 2187, so n – 1 = 7 n = 8
So 4374 is the 8th term. Check: 2 × 37 = 2 × 2187 = 4374. ✓
Why it happens: in a GP the unknown position sits in the exponent, so the last step is not a division but a matching of powers: you must recognise 2187 as 37. Build the powers of 3 up from the bottom — 3, 9, 27, 81, 243, 729, 2187 — and count them. If 4374 had turned out not to be 2 times a power of 3, it would simply not be a term of this GP.
Q5.
A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60% of the height from which it fell. It continues bouncing in this way — each time rising to 60% of the previous height. (i) What height does the ball reach after the 5th bounce? (ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?
Answer
The peak heights form a GP with r = 60% = 0.6, starting from the 80 m drop.
After 1st bounce: 80 × 0.6 = 48 m After 2nd bounce: 48 × 0.6 = 28.8 m After 3rd bounce: 28.8 × 0.6 = 17.28 m After 4th bounce: 17.28 × 0.6 = 10.368 m After 5th bounce: 10.368 × 0.6 = 6.2208 m
(i) The height after the 5th bounce is 6.2208 m (about 6.22 m). Directly: 80 × (0.6)5 = 80 × 0.07776 = 6.2208 m.
(ii) Count the journeys carefully. The ball falls 80 m, then each bounce is a rise followed by an equal fall, and the 6th hit on the ground happens after the fall that follows the 5th bounce.
Why it happens: the heights make a GP, but the distance travelled is not a single term of it — it is a total, and every rise after the first fall is matched by a fall of the same length. That is why the five peak heights are doubled while the original 80 m is counted once. Miscounting here is the usual error: the 6th hit comes after only 5 bounces.
Tip: 60% means r = 0.6, so the ball loses 40% of its height each time. After 5 bounces only (0.6)5 ≈ 7.8% of the original height is left.
Q6.
Which term of the sequence 2, 2√2, 4, … is 128?
Answer
First check that this is a GP by taking ratios.
2√2 ÷ 2 = √2 4 ÷ 2√2 = 2/√2 = √2 Constant, so a = 2 and r = √2
Write every term as a power of 2, using √2 = 21/2:
tn = 2 × (√2)n–1 = 21 × 2(n–1)/2 = 2(n+1)/2 128 = 27 So (n + 1)/2 = 7 n + 1 = 14 n = 13
Why it happens: the sequence is 2, 2√2, 4, 4√2, 8, 8√2, 16, … — the whole numbers double only at every second step, because two multiplications by √2 make one multiplication by 2. Rewriting the surd as the power 21/2 turns an awkward-looking sequence into plain powers of 2, and then matching exponents finishes the job in one line.
Check it yourself: the terms with whole-number values sit at the odd positions — t1 = 2, t3 = 4, t5 = 8, t7 = 16, t9 = 32, t11 = 64, t13 = 128.
Q7.
Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 of this fractal is a square sheet of paper. To construct Stage 1, each side of the square is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares. The centre square is then removed and the 8 smaller squares are retained, leaving a square hole in the centre. The same process is repeated on the eight smaller shaded squares to obtain Stage 2 and so on. Look at Fig. 8.12 and try to answer the following questions. (i) How many red squares are there in Stages 0 to 3? (ii) Can you predict the number of red squares in Stages 4 and 5? (iii) Can you find a rule for the number of red squares at the nth stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage. (iv) Suppose the area of the square in Stage 0 is 1 square unit. What is the area of the red region in Stages 1, 2 and 3? What will be the area of the red region in Stages 4 and 5? Find the explicit as well as the recursive formula for the area of the red region at the nth stage. What happens to this area as n, the number of stages, goes on increasing?
Fig. 8.12 — Stages 0, 1, 2 and 3 of the Sierpiński square carpet.
Answer
Every red square becomes 9 smaller squares of which 8 are kept, so both counts and areas are governed by the same step.
One step of the construction: 9 equal squares, the centre one removed, 8 kept — so the count is multiplied by 8 and the area by 8/9.
(iii) The counts 1, 8, 64, 512, … are a GP with common ratio 8, and they are the powers of 8 with exponent equal to the stage number.
Explicit: tn = 8n Recursive: t0 = 1, tn = 8 × tn–1 for n ≥ 1
(iv) Each of the nine small squares has 1/9 of the area of the square it came from, and 8 of them are kept, so the red area is multiplied by 8/9 at every stage.
Since 8/9 is less than 1, the area falls at every stage and approaches 0 as n grows — though it never actually reaches 0.
Why it happens: this is the Sierpiński triangle's argument with different numbers. Trisecting the sides makes 3 × 3 = 9 congruent squares, so each has area 1/9 of the original; keeping 8 of them leaves 8 × (1/9) = 8/9 of the area but multiplies the number of pieces by 8. Because 8/9 is closer to 1 than the triangle's 3/4, the carpet's area shrinks more slowly — but it shrinks all the same, while 8n outruns 3n. A fractal is exactly this: more and more pieces, less and less area.
Check it yourself: at Stage n there are 8n red squares, each of area (1/9)n. Their total is 8n × (1/9)n = (8/9)n ✓ — the two formulas fit together.