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NCERT Class 9 Mathematics Exercise Set 8.3 Solutions: Predicting What Comes Next: Exploring Sequences and Progressions

Step-by-step working for every question of Exercise Set 8.3 in NCERT's Class 9 Mathematics book, chapter “Predicting What Comes Next: Exploring Sequences and Progressions”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Ex 8.3 Q1

   A GP has common ratio 2 and 8th term 192. Find its 12th term.

   1. The formula for the nth term of a GP is aₙ = a × rⁿ⁻¹.
      why?

      In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.

   2. Given the 8th term a₈ = 192 and common ratio r = 2.
   3. Substitute these values into the formula: 192 = a × 2⁷.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   4. Calculate 2⁷ which is 128, so 192 = a × 128.
   5. Solve for the first term a: a = 192 ÷ 128 = 1.5.
   6. Now find the 12th term a₁₂ using a = 1.5 and r = 2.
   7. a₁₂ = 1.5 × 2¹¹.
   8. Calculate 2¹¹ which is 2048.
   9. Multiply 1.5 by 2048 to get 3072.

   Answer: 3072

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2. Ex 8.3 Q2

   10th term and nth term of the GP 5, 25, 125, …

   1. The first term is a = 5 and the common ratio is r = 25 ÷ 5 = 5 (also 125 ÷ 25 = 5).
   2. The nth term of a GP is tₙ = a × rⁿ⁻¹, so tₙ = 5 × 5ⁿ⁻¹ = 5ⁿ.
      why?

      In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.

   3. Check: n = 1 gives 5, n = 2 gives 25, n = 3 gives 125, matching the given terms.
   4. The 10th term is t₁₀ = 5¹⁰.
   5. 5¹⁰ = 5⁵ × 5⁵ = 3125 × 3125 = 9765625.

   Answer: t₁₀ = 9765625; tₙ = 5ⁿ

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3. Ex 8.3 Q3

   t₁ = 2, tₙ₊₁ = 3tₙ − 2 (n ≥ 1). Which term is 730?

   1. Write the recurrence relation as tₙ₊₁ = 3tₙ − 2 with t₁ = 2.
   2. Calculate the 2nd term: t₂ = 3(2) − 2 = 4.
   3. Calculate the 3rd term: t₃ = 3(4) − 2 = 10.
   4. Calculate the 4th term: t₄ = 3(10) − 2 = 28.
   5. Calculate the 5th term: t₅ = 3(28) − 2 = 82.
   6. Calculate the 6th term: t₆ = 3(82) − 2 = 244.
   7. Calculate the 7th term: t₇ = 3(244) − 2 = 730.

   Answer: 7th term

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4. Ex 8.3 Q4

   GP 2, 6, 18, …: which term is 4374? Give its explicit and recursive formulas.

   1. Identify the first term a = 2 and the common ratio r = 6 ÷ 2 = 3.
   2. Use the general term formula for a GP: tₙ = a × rⁿ⁻¹.
   3. Substitute the known values to get tₙ = 2 × 3ⁿ⁻¹.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   4. Set the general term equal to the given value: 2 × 3ⁿ⁻¹ = 4374.
   5. Divide both sides by 2 to isolate the power of 3: 3ⁿ⁻¹ = 2187.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   6. Express 2187 as a power of 3: 2187 = 3⁷.
   7. Equate the exponents to find n − 1 = 7.
   8. Solve for n to get n = 8.
   9. The explicit formula is tₙ = 2 × 3ⁿ⁻¹.
   10. The recursive formula is tₙ = 3tₙ₋₁ with t₁ = 2.

   Answer: 8th term; tₙ = 2 × 3ⁿ⁻¹; tₙ = 3tₙ₋₁

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5. Ex 8.3 Q5

   A ball dropped from 80 m bounces back each time to 60% of the height it fell from. (i) Height reached after the 5th bounce? (ii) Total vertical distance travelled when it hits the ground for the 6th time?

   1. (i) The initial height is 80 m and the ball bounces to 60% of the previous height.
      why?

      Per cent means 'out of 100', so x% of a quantity = (x / 100) × the quantity.

   2. (i) The height after the 1st bounce is 80 × 0.6 = 48 m.
   3. (i) The height after the 2nd bounce is 48 × 0.6 = 28.8 m.
   4. (i) The height after the 3rd bounce is 28.8 × 0.6 = 17.28 m.
   5. (i) The height after the 4th bounce is 17.28 × 0.6 = 10.368 m.
   6. (i) The height after the 5th bounce is 10.368 × 0.6 = 6.2208 m.
   7. (ii) The total distance includes the initial fall of 80 m.
   8. (ii) The ball travels up and down for the first 5 bounces before hitting the ground for the 6th time.
   9. (ii) The sum of the heights for the first 5 bounces is 48 + 28.8 + 17.28 + 10.368 + 6.2208 = 110.6688 m.
   10. (ii) The total vertical distance is 80 + 2 × 110.6688 = 301.3376 m.

   Answer: (i) 6.2208 m; (ii) 301.3376 m

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6. Ex 8.3 Q6

   Which term of 2, 2√2, 4, … is 128?

   1. The given sequence is 2, 2√2, 4, ...
   2. Calculate the common ratio r by dividing the second term by the first term: r = 2√2 ÷ 2 = √2.
   3. Verify the ratio with the third term: 2√2 × √2 = 2 × 2 = 4, which matches the given term.
   4. Use the formula for the nth term of a geometric progression: aₙ = a × rⁿ⁻¹.
      why?

      In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.

   5. Substitute the known values into the formula: 128 = 2 × (√2)ⁿ⁻¹.
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   6. Divide both sides by 2 to isolate the exponential term: 64 = (√2)ⁿ⁻¹.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   7. Express 64 as a power of √2. Since (√2)² = 2, then (√2)⁶ = 2³ = 8 and (√2)¹² = 2⁶ = 64.
   8. Equate the exponents: n − 1 = 12.
   9. Solve for n: n = 12 + 1 = 13.

   Answer: 13th term

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7. Ex 8.3 Q7

   Sierpiński square carpet (Fig. 8.12): Stage 0 is a square; each stage cuts every red square into 9 and removes the centre one. (i) Red squares in Stages 0 to 3; (ii) in Stages 4 and 5; (iii) explicit and recursive rule for Stage n; (iv) red area in Stages 1 to 5 if Stage 0 has area 1, explicit and recursive formulas, and what happens as n grows.

   Uses a figure printed in the NCERT book — open the chapter above to see it.

   1. (i) Stage 0 has 1 red square.
   2. (i) Stage 1 has 8 red squares.
   3. (i) Stage 2 has 64 red squares.
   4. (i) Stage 3 has 512 red squares.
   5. (ii) Stage 4 has 4096 red squares.
   6. (ii) Stage 5 has 32768 red squares.
   7. (iii) The explicit rule for the number of red squares at Stage n is 8ⁿ.
   8. (iii) The recursive rule is aₙ = 8 × aₙ₋₁ with a₀ = 1.
      why?

      In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.

   9. (iv) The red area at Stage 1 is 8/9.
   10. (iv) The red area at Stage 2 is 64/81.
   11. (iv) The red area at Stage 3 is 512/729.
   12. (iv) The red area at Stage 4 is 4096/6561.
   13. (iv) The red area at Stage 5 is 32768/59049.
   14. (iv) The explicit formula for red area at Stage n is (8/9)ⁿ.
   15. (iv) The recursive formula is Aₙ = (8/9) × Aₙ₋₁ with A₀ = 1.
       why?

       In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.

   16. (iv) As n grows, the red area approaches 0.

   Answer: (i) 1, 8, 64, 512; (ii) 4096, 32768; (iii) 8ⁿ, aₙ = 8aₙ₋₁; (iv) 8/9, 64/81, 512/729, 4096/6561, 32768/59049, (8/9)ⁿ, Aₙ = (8/9)Aₙ₋₁, approaches 0

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