Home · Schooling · CBSE · Class 9 · Mathematics · Predicting What Comes Next: Exploring Sequences and Progressions · End-of-chapter exercises

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

Step-by-step working for every question of End-of-chapter exercises 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. Exercises Q1

   An AP has 11th term 38 and 16th term 73. Find its 31st term.

   1. Let the first term be a and the common difference be d.
      why?

      In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.

   2. The 11th term is given by a + 10d = 38.
   3. The 16th term is given by a + 15d = 73.
   4. Subtracting the first equation from the second gives 5d = 35.
   5. Solving for d gives d = 7.
   6. Substitute d = 7 into a + 10d = 38 to get a + 70 = 38.
      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.

   7. Solving for a gives a = -32.
   8. The 31st term is given by a + 30d.
   9. Substitute the values to get -32 + 30(7) = -32 + 210.
      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.

   10. The result is 178.

   Answer: 178

   Report an error in this solution

2. Exercises Q2

   Find the AP whose 3rd term is 16 and whose 7th term is 12 more than its 5th term.

   1. Let the first term be a and the common difference be d.
      why?

      In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.

   2. The 3rd term is given as 16, so a + 2d = 16.
   3. The 7th term is 12 more than the 5th term, so a + 6d = (a + 4d) + 12.
   4. Simplifying the second equation gives 2d = 12, so d = 6.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   5. Substitute d = 6 into a + 2d = 16 to get a + 12 = 16.
      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. Solving for a gives a = 4.
   7. The first few terms are 4, 10, 16, 22.

   Answer: 4, 10, 16, 22

   Report an error in this solution

3. Exercises Q3

   How many three-digit numbers are divisible by 7?

   1. The smallest three-digit number divisible by 7 is 105.
   2. The largest three-digit number divisible by 7 is 994.
   3. These numbers form an arithmetic progression with first term a = 105, common difference d = 7, and last term l = 994.
      why?

      In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.

   4. Using the formula for the nth term, l = a + (n − 1)d.
      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. Substituting the values gives 994 = 105 + (n − 1)7.
      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. Simplifying, 994 − 105 = 7(n − 1).
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   7. 889 = 7(n − 1).
   8. Dividing both sides by 7, n − 1 = 127.
      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.

   9. Adding 1 to both sides, n = 128.
      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.

   Answer: 128

   Report an error in this solution

4. Exercises Q4

   How many multiples of 4 lie between 10 and 250?

   1. The first multiple of 4 greater than 10 is 12.
   2. The last multiple of 4 less than 250 is 248.
   3. This forms an arithmetic progression with first term a = 12 and common difference d = 4.
      why?

      In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.

   4. Using the formula for the nth term, 248 = 12 + (n − 1) × 4.
      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. Subtracting 12 from both sides gives 236 = (n − 1) × 4.
      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. Dividing by 4 gives n − 1 = 59.
   7. Adding 1 to both sides gives n = 60.
      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.

   Answer: 60

   Report an error in this solution

5. Exercises Q5

   Find a GP whose first two terms add to −4 and whose 5th term is 4 times its 3rd term.

   1. Let the first term be a and the common ratio be r.
   2. The sum of the first two terms is a + ar, so a(1 + r) = −4.
      why?

      Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.

   3. The 5th term is 4 times the 3rd term, so ar⁴ = 4ar².
   4. Dividing by ar² (since a and r are non-zero), we get r² = 4.
   5. Solving for r gives r = 2 or r = −2.
   6. If r = 2, then a(1 + 2) = −4, so 3a = −4 and a = −4/3.
   7. The first three terms are −4/3, −8/3, and −16/3.
   8. If r = −2, then a(1 − 2) = −4, so −a = −4 and a = 4.
   9. The first three terms are 4, −8, and 16.

   Answer: The two possible geometric progressions are −4/3, −8/3, −16/3, ... and 4, −8, 16, ...

   Report an error in this solution

6. Exercises Q6

   Find all ways of writing 100 as a sum of consecutive natural numbers.

   1. Let the sum of n consecutive natural numbers starting from a be 100.
   2. The sum is given by the formula n/2 × (2a + n − 1) = 100.
   3. Simplifying, we get n(2a + n − 1) = 200.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   4. Since a is a natural number, a ≥ 1, so 2a + n − 1 ≥ n, which means n² ≤ 200.
   5. Thus, n must be a divisor of 200 such that n² ≤ 200, so n can be 1, 2, 4, 5, 8, or 10.
   6. For n = 1, 2a = 200 gives a = 100: that is just the number 100 itself, not a sum of two or more consecutive numbers.
   7. For n = 2, 2a + 1 = 100, so 2a = 99, which gives a non-integer value.
   8. For n = 4, 2a + 3 = 50, so 2a = 47, which gives a non-integer value.
   9. For n = 5, 2a + 4 = 40, so 2a = 36, giving a = 18.
   10. The sequence for n = 5 is 18 + 19 + 20 + 21 + 22.
   11. For n = 8, 2a + 7 = 25, so 2a = 18, giving a = 9.
   12. The sequence for n = 8 is 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16.
   13. For n = 10, 2a + 9 = 20, so 2a = 11, which gives a non-integer value.
   14. Therefore, only two ways exist to write 100 as a sum of consecutive natural numbers.

   Answer: 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16; 18 + 19 + 20 + 21 + 22

   Report an error in this solution

7. Exercises Q7

   A culture starts with 30 bacteria and doubles every hour. Number at the end of hour 2, hour 4 and hour n?

   1. The initial number of bacteria is 30.
   2. The bacteria double every hour, so the number after n hours is 30 × 2ⁿ.
   3. For hour 2, substitute n = 2 into the formula: 30 × 2².
      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 30 × 4 to get 120.
   5. For hour 4, substitute n = 4 into the formula: 30 × 2⁴.
      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. Calculate 30 × 16 to get 480.
   7. For hour n, the general expression is 30 × 2ⁿ.

   Answer: 120; 480; 30 × 2ⁿ

   Report an error in this solution

8. Exercises Q8

   In an AP, t₄ + t₈ = 24 and t₆ + t₁₀ = 44. Find the first three terms.

   1. Let the first term be a and the common difference be d.
      why?

      In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.

   2. The 4th term is a + 3d and the 8th term is a + 7d, so their sum is 2a + 10d = 24.
   3. The 6th term is a + 5d and the 10th term is a + 9d, so their sum is 2a + 14d = 44.
   4. Subtracting the first equation from the second gives 4d = 20, so d = 5.
   5. Substitute d = 5 into 2a + 10d = 24 to get 2a + 50 = 24.
      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. Solving for a gives 2a = −26, so a = −13.
   7. The first term is −13, the second term is −13 + 5 = −8, and the third term is −8 + 5 = −3.

   Answer: −13, −8, −3

   Report an error in this solution

9. Exercises Q9

   Smallest n for which 1 + 2 + … + n is greater than 1,000.

   1. The sum of the first n natural numbers is given by the formula S = n(n + 1) / 2.
   2. We need to find the smallest n such that S > 1000.
   3. Substitute the formula into the inequality: n(n + 1) / 2 > 1000.
      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. Multiply both sides by 2 to get n(n + 1) > 2000.
      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.

   5. Test n = 44: 44 × 45 = 1980, which is less than 2000.
   6. Test n = 45: 45 × 46 = 2070, which is greater than 2000.
   7. Since 2070 > 2000, the sum for n = 45 is greater than 1000.
   8. Therefore, the smallest such value of n is 45.

   Answer: n = 45

   Report an error in this solution

10. Exercises Q10

    GP 2, 8, 32, …: which term is 131072? Give its explicit and recursive formulas.

    1. Identify the first term a = 2 and the common ratio r = 8 ÷ 2 = 4.
    2. Use the general term formula for a GP: tₙ = a × rⁿ⁻¹.
    3. Substitute the given term 131072: 131072 = 2 × 4ⁿ⁻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. Divide both sides by 2: 65536 = 4ⁿ⁻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.

    5. Express 65536 as a power of 4: 4⁸ = 65536.
    6. Equate the exponents: n − 1 = 8.
    7. Solve for n: n = 9.
    8. The explicit formula is tₙ = 2 × 4ⁿ⁻¹.
    9. The recursive formula is tₙ = 4tₙ₋₁.

    Answer: 9th term; tₙ = 2 × 4ⁿ⁻¹; tₙ = 4tₙ₋₁

    Report an error in this solution

11. Exercises Q11

    The first three terms of a GP add to 13/12 and multiply to −1. Find the common ratio and the terms.

    1. Let the three terms of the GP be a/r, a, and ar.
    2. The product of the terms is (a/r) × a × ar = a³.
    3. Given the product is −1, we have a³ = −1, so a = −1.
    4. The sum of the terms is a/r + a + ar = 13/12.
    5. Substitute a = −1 into the sum equation: −1/r − 1 + (−1)r = 13/12.
       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. Simplify to −1/r − 1 − r = 13/12.
       why?

       Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

    7. Multiply the entire equation by 12r to clear denominators: −12 − 12r − 12r² = 13r.
    8. Rearrange into a standard quadratic form: 12r² + 25r + 12 = 0.
    9. Factor the quadratic: (3r + 4)(4r + 3) = 0.
    10. Solving for r gives r = −4/3 or r = −3/4.
    11. If r = −4/3, the terms are a/r = (−1)/(−4/3) = 3/4, a = −1, and ar = (−1)(−4/3) = 4/3.
    12. If r = −3/4, the terms are a/r = (−1)/(−3/4) = 4/3, a = −1, and ar = (−1)(−3/4) = 3/4.
    13. Both sets of terms are valid solutions for the GP.

    Answer: Common ratio: −3/4 or −4/3; Terms: 4/3, −1, 3/4 or 3/4, −1, 4/3

    Report an error in this solution

12. Exercises Q12

    The 4th, 10th and 16th terms of a GP are x, y, z. Prove that x, y, z are in GP.

    1. Let the first term of the given GP be a and the common ratio be r.
    2. The 4th term is x = ar³, the 10th term is y = ar⁹, and the 16th term is z = ar¹⁵.
    3. To prove that x, y, z are in GP, we need to show that y² = xz.
    4. Substitute the expressions for x, y, and z into the equation y² = xz.
       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.

    5. The left side is (ar⁹)² = a²r¹⁸.
    6. The right side is (ar³)(ar¹⁵) = a²r¹⁸.
    7. Since both sides are equal, y² = xz holds true.
       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.

    8. Therefore, x, y, z are in geometric progression.

    Answer: Proved that x, y, z are in GP.

    Report an error in this solution

13. Exercises Q13

    Three terms of a GP add to 26 and their squares add to 364. Find the terms.

    1. Let the three terms of the GP be a/r, a, and ar.
    2. The sum of the terms is given as 26, so a/r + a + ar = 26.
    3. Factor out a to get a(1/r + 1 + r) = 26.
    4. The sum of the squares is given as 364, so (a/r)² + a² + (ar)² = 364.
    5. Factor out a² to get a²(1/r² + 1 + r²) = 364.
    6. Divide the square sum equation by the linear sum equation: [a²(1/r² + 1 + r²)] / [a(1/r + 1 + r)] = 364 / 26.
    7. Simplify the left side to a(1/r² + 1 + r²) / (1/r + 1 + r) and the right side to 14.
       why?

       Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

    8. Let S = 1/r + 1 + r. Then 1/r² + 1 + r² = S² − 2(1 + r + 1/r) = S² − 2S.
    9. Substitute into the ratio: a(S² − 2S) / S = 14, which simplifies to a(S − 2) = 14.
       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.

    10. From the sum equation, aS = 26, so a = 26/S.
    11. Substitute a into the previous equation: (26/S)(S − 2) = 14.
        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.

    12. Multiply by S: 26(S − 2) = 14S.
    13. Expand: 26S − 52 = 14S.
        why?

        Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

    14. Rearrange: 12S = 52, so S = 52/12 = 13/3.
    15. Find a: a = 26 / (13/3) = 26 × 3/13 = 6.
    16. Find r using S = 1/r + 1 + r = 13/3.
    17. Subtract 1: 1/r + r = 13/3 − 1 = 10/3.
    18. Multiply by r: 1 + r² = (10/3)r.
    19. Rearrange: 3r² − 10r + 3 = 0.
    20. Factor: (3r − 1)(r − 3) = 0.
    21. So r = 3 or r = 1/3.
    22. If r = 3, the terms are 6/3, 6, 6×3, which are 2, 6, 18.
    23. If r = 1/3, the terms are 6/(1/3), 6, 6×(1/3), which are 18, 6, 2.
    24. Both sets are the same terms in different order.

    Answer: 2, 6, 18

    Report an error in this solution

14. Exercises Q14

    P₁ = 1, P₂ = 2 and Pₙ = P₁ + P₂ + … + Pₙ₋₁ + 1 for n > 2. Find P₁ to P₈, a simpler recursive rule and an explicit formula.

    1. Given P₁ = 1 and P₂ = 2.
    2. For n > 2, Pₙ is the sum of all previous terms plus 1.
    3. Calculate P₃ = P₁ + P₂ + 1 = 1 + 2 + 1 = 4.
    4. Calculate P₄ = P₁ + P₂ + P₃ + 1 = 1 + 2 + 4 + 1 = 8.
    5. Calculate P₅ = P₁ + P₂ + P₃ + P₄ + 1 = 1 + 2 + 4 + 8 + 1 = 16.
    6. Calculate P₆ = P₁ + P₂ + P₃ + P₄ + P₅ + 1 = 1 + 2 + 4 + 8 + 16 + 1 = 32.
    7. Calculate P₇ = P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + 1 = 1 + 2 + 4 + 8 + 16 + 32 + 1 = 64.
    8. Calculate P₈ = P₁ + P₂ + P₃ + P₄ + P₅ + P₆ + P₇ + 1 = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 1 = 128.
    9. The sequence is 1, 2, 4, 8, 16, 32, 64, 128.
    10. For n ≥ 4, Pₙ = (P₁ + … + Pₙ₋₂ + 1) + Pₙ₋₁ = Pₙ₋₁ + Pₙ₋₁, because the bracket is exactly Pₙ₋₁; and P₃ = 4 = 2P₂. So the simpler recursive rule is Pₙ = 2Pₙ₋₁ (n ≥ 2, since P₂ = 2 = 2P₁ too).
    11. Since P₁ = 1 and the common ratio is 2, the explicit formula is Pₙ = 2ⁿ⁻¹.

    Answer: P₁ to P₈ are 1, 2, 4, 8, 16, 32, 64, 128; the simpler recursive rule is Pₙ = 2Pₙ₋₁; the explicit formula is Pₙ = 2ⁿ⁻¹

    Report an error in this solution

15. Exercises Q15

    W₁ = 1, W₂ = 2 and Wₙ = W₁ + W₂ + … + Wₙ₋₂ + 2 for n > 2. Find W₁ to W₈. Which known sequence is it?

    1. Given W₁ = 1 and W₂ = 2.
    2. For n > 2, the rule is Wₙ = W₁ + W₂ + … + Wₙ₋₂ + 2.
    3. Calculate W₃: the rule adds the terms up to W₁, plus 2, so W₃ = W₁ + 2 = 1 + 2 = 3.
    4. Calculate W₄: the terms up to W₂, plus 2, so W₄ = 1 + 2 + 2 = 5.
    5. Calculate W₅: sum of terms up to W₃ plus 2, so W₅ = 1 + 2 + 3 + 2 = 8.
    6. Calculate W₆: sum of terms up to W₄ plus 2, so W₆ = 1 + 2 + 3 + 5 + 2 = 13.
    7. Calculate W₇: sum of terms up to W₅ plus 2, so W₇ = 1 + 2 + 3 + 5 + 8 + 2 = 21.
    8. Calculate W₈: sum of terms up to W₆ plus 2, so W₈ = 1 + 2 + 3 + 5 + 8 + 13 + 2 = 34.
    9. The sequence 1, 2, 3, 5, 8, 13, 21, 34 follows the pattern where each term is the sum of the two preceding terms.
    10. This is the Virahānka–Fibonacci sequence.

    Answer: 1, 2, 3, 5, 8, 13, 21, 34; Virahānka–Fibonacci

    Report an error in this solution

← Predicting What Comes Next: Exploring Sequences and Progressions: notes, practice and all exercises