Class 9 CBSE — Geometric Progression Practice Sheet

🧠 30-Second GP Recall

GP:
Each term is obtained by multiplying the previous term by the same non-zero number.

Common ratio:
r = second term ÷ first term

n-th term from beginning:
a_n = arⁿ⁻¹

n-th term from end:
If last term is l, T_n from end = lr⁻⁽ⁿ⁻¹⁾

Middle term:
For an odd number of terms, the middle term is T_(N+1)/2.

Two middle terms:
For an even number of terms, positions are N/2 and N/2+1.

Memory trick: “GP = × same number.”
To move forward, multiply by r. To move backward, divide by r.

📘 Additional Practice — Selected Questions

Each solution is hidden. Click Show Full Answer after attempting the question.

2. Find the 10^th and n^th terms of the GP: 5, 25, 125, ...

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Step 1: Identify the first term and common ratio.

a = 5, r = 25/5 = 5

Step 2: Use the n-th term formula.

T_n = arⁿ⁻¹

Therefore,

T_n = 5(5)ⁿ⁻¹ = 5ⁿ

10th term:

T₁₀ = 5¹⁰ = 9,765,625

Answer: 10th term = 9,765,625; n-th term = 5ⁿ.

3. A sequence is given by the recursive rule t₁ = 2, t_n+1 = 3t_n − 2 for n ≥ 1. Which term of the sequence is 730?

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Calculate a few terms first:

t₁=2

t₂=3(2)−2=4

t₃=3(4)−2=10

t₄=28

t₅=82

t₆=244

t₇=730

So 730 is the 7th term.

Shortcut: Put u_n=t_n−1. Then

u_n+1=3u_n, u₁=1

Hence u_n=3ⁿ⁻¹, so

t_n=1+3ⁿ⁻¹

For 730: 730=1+3⁶, hence n=7.

4. Which term of the GP 2, 6, 18, ... is 4374? Write the explicit formula for the n-th term as well as the recursive formula.

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Here a=2 and r=3.

Explicit formula:

T_n=2(3)ⁿ⁻¹

For the required term:

2(3)ⁿ⁻¹=4374

3ⁿ⁻¹=2187=3⁷

Therefore n−1=7, so n=8.

Recursive formula:

T₁=2, T_n+1=3T_n

Answer: 4374 is the 8th term.

5. 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 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?

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(i) Height after the 5th bounce

The bounce heights form a GP:

48, 48(0.6), 48(0.6)², ...

So the 5th bounce height is

H₅=80(0.6)⁵

=80(0.07776)=6.2208 m

Answer: 6.2208 m.

(ii) Total distance up to the 6th ground hit

The ball first travels 80 m downward. Then it travels up and down through the first five bounce heights.

Thus

D=80+2[48+28.8+17.28+10.368+6.2208]

The five bounce heights form a GP with a=48, r=0.6, n=5.

S₅=48(1−0.6⁵)/(1−0.6)=110.6688

Therefore

D=80+2(110.6688)=301.3376 m

Answer: Total vertical distance = 301.3376 m.

6. Which term of the sequence 2, 2√2, 4, ... is 128?

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The common ratio is

r=(2√2)/2=√2

Using T_n=arⁿ⁻¹:

128=2(√2)ⁿ⁻¹

64=(√2)ⁿ⁻¹

Since 64=2⁶ and √2=2^1/2,

2⁶=2^(n−1)/2

Therefore 6=(n−1)/2, so n−1=12.

Answer: 128 is the 13th term.

7. The figure shows Stages 0 to 3 of the Sierpiński square carpet. At each stage, each red square is divided into 9 equal squares and the centre square is removed, leaving 8 smaller red squares.

(i) How many red squares are there in Stages 0 to 3?
(ii) Predict the number of red squares in Stages 4 and 5.
(iii) Find an explicit formula and a recursive formula for the number of red squares at the n-th stage.
(iv) If the area of the square in Stage 0 is 1 square unit, find the area of the red region in Stages 1, 2 and 3. Find the area in Stages 4 and 5. Find explicit and recursive formulas for the area of the red region at the n-th stage. What happens to this area as the number of stages increases?

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(i) Number of red squares

Stage 0: 1

Stage 1: 8

Stage 2: 8×8=64

Stage 3: 8×8×8=512

Answer: 1, 8, 64, 512.

(ii) Stages 4 and 5

Stage 4=8⁴=4096

Stage 5=8⁵=32768

(iii) Formula for number of red squares

Each stage multiplies the number of red squares by 8. Therefore it is a GP with first term 1 and ratio 8.

R_n=8ⁿ

Recursive form:

R₀=1, R_n=8R_n−1

(iv) Area of the red region

At each stage, every red square keeps 8/9 of its area. Therefore the areas form a GP:

Stage 0: 1

Stage 1: 8/9

Stage 2: 64/81

Stage 3: 512/729

Stage 4: 4096/6561

Stage 5: 32768/59049

The explicit formula is

A_n=(8/9)ⁿ

Recursive formula:

A₀=1, A_n=(8/9)A_n−1

Because 0<8/9<1, the area keeps decreasing and approaches 0 square units as the number of stages becomes very large.

A. Find the Common Ratio

1. Find r for: 3, 12, 48, 192, ...

2. Find r for: 81, 27, 9, 3, ...

3. Find r for: −2, 6, −18, 54, ...

4. Is 2, 6, 18, 54, 162 a GP? Give a reason.

5. Is 5, 10, 20, 41, 82 a GP? Explain without finding all terms.

B. n-th Term from the Beginning

6. Find the 8th term of 2, 6, 18, 54, ...

7. Find the 10th term of 5, 10, 20, 40, ...

8. Find the 7th term of 128, 64, 32, 16, ...

9. The first term of a GP is 3 and common ratio is 2. Find its 12th term.

10. The first term is 64 and common ratio is 1/2. Which term is 4?

C. n-th Term from the End

Recall: If the last term is l, move backward by dividing by r each time.

11. Find the 3rd term from the end of 2, 6, 18, 54, 162.

12. Find the 4th term from the end of 3, 9, 27, 81, 243, 729.

13. A GP has 8 terms. Its first term is 2 and common ratio is 3. Find the 5th term from the end.

14. A GP has 10 terms. Its last term is 1536 and common ratio is 2. Find the 6th term from the end.

D. Middle Term(s)

15. Find the middle term of: 2, 6, 18, 54, 162, 486, 1458.

16. Find the middle term of a GP having 9 terms, first term 3 and common ratio 2.

17. A GP has 11 terms. Which term is the middle term? Then write its position from the end.

18. A GP has 12 terms. Are there one or two middle terms? Name their positions from the beginning.

19. Find both middle terms of 2, 4, 8, 16, 32, 64, 128, 256.

E. 🔥 HOT Questions — Think, Don't Just Apply Formula

20. Three consecutive terms of a GP are x, 12, 48. Find x.

21. Three consecutive terms of a GP are x−2, x, x+8. Find x.

22. The 3rd term of a GP is 12 and the 6th term is 96. Find the common ratio, assuming it is positive.

23. The 4th term of a GP is 24 and the 7th term is 192. Find the first term and common ratio.

24. A GP has first term 5 and common ratio 3. One of its terms is 405. Which term is it?

25. In a GP, the 5th term is 81 and the 8th term is 2187. Find r and the first term.

26. A GP has 9 terms. The first term is 2 and the last term is 512. Find the middle term without writing all the terms.

27. The 4th term from the beginning is equal to the 4th term from the end of a GP with 7 terms. What does this tell you about the sequence?

F. MCQ Revision

28. Which is a GP?

(A) 2, 4, 8, 16(B) 2, 4, 7, 11(C) 3, 6, 10, 15(D) 1, 3, 6, 10

29. The common ratio of 81, 27, 9, 3 is:

(A) 3(B) 1/3(C) −3(D) −1/3

30. The 6th term of 2, 4, 8, ... is:

(A) 16(B) 32(C) 64(D) 128

31. A GP has 15 terms. Its middle term is:

(A) 7th(B) 8th(C) 7th and 8th(D) 15th

32. The 3rd term from the end of 3, 6, 12, 24, 48 is:

(A) 6(B) 12(C) 24(D) 48

33. If a=5, r=−2, then the 4th term is:

(A) −40(B) 40(C) −20(D) 20

34. If the first term is 4 and the common ratio is 3, which term is 108?

(A) 3rd(B) 4th(C) 5th(D) 6th

35. If three consecutive terms of a GP are x/r, x, xr, their product is:

(A) x(B) x²(C) x³(D) r³

36. For a GP with 9 terms, the 4th term from the beginning and 6th term from the end are:

(A) same term(B) adjacent terms(C) first and last(D) cannot be related

G. Final Recall Challenge

Q1. What is the fastest way to test whether a sequence is GP?

Q2. How do you move one position toward the beginning?

Q3. For 13 terms, which term is in the middle?

Q4. For 14 terms, how many middle terms are there?

🔒 Teacher Answer Key — hidden from students

1. 4

2. 1/3

3. −3

4. Yes, ratio = 3.

5. No; 41/20 ≠ 2.

6. 4374

7. 2560

8. 2

9. 6144

10. 5th term

11. 54

12. 27

13. 54

14. 48

15. 54

16. 48

17. 6th term; also 6th from end.

18. Two middle terms: 6th and 7th.

19. 16 and 32.

20. 3

21. x = 4

22. r = 2; first term = 3.

23. r = 2; first term = 3.

24. 5th term.

25. r = 3; first term = 1.

26. 32.

27. The 4th term is the middle term; the sequence has 7 terms and is symmetric in position about the middle.

28. A 29. B 30. B 31. B 32. B 33. A 34. B 35. C 36. A

Final Recall: Q1: Divide consecutive terms and check the ratio is constant. Q2: Divide by r. Q3: 7th. Q4: 2.

Teacher use: keep the answer key collapsed when projecting the sheet. Students can complete the questions without seeing the solutions.