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NCERT Class 9 Mathematics Exercise Set 8.1 Solutions: Predicting What Comes Next: Exploring Sequences and Progressions

Step-by-step working for every question of Exercise Set 8.1 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 8.1 Q1

   First five terms (n ≥ 1) of (i) tₙ = 3n − 4, (ii) tₙ = 2 − 5n, (iii) tₙ = n² − 2n + 3.

   1. (i) For n = 1, t₁ = 3(1) − 4 = −1.
   2. (i) For n = 2, t₂ = 3(2) − 4 = 2.
   3. (i) For n = 3, t₃ = 3(3) − 4 = 5.
   4. (i) For n = 4, t₄ = 3(4) − 4 = 8.
   5. (i) For n = 5, t₅ = 3(5) − 4 = 11.
   6. (ii) For n = 1, t₁ = 2 − 5(1) = −3.
   7. (ii) For n = 2, t₂ = 2 − 5(2) = −8.
   8. (ii) For n = 3, t₃ = 2 − 5(3) = −13.
   9. (ii) For n = 4, t₄ = 2 − 5(4) = −18.
   10. (ii) For n = 5, t₅ = 2 − 5(5) = −23.
   11. (iii) For n = 1, t₁ = 1² − 2(1) + 3 = 2.
   12. (iii) For n = 2, t₂ = 2² − 2(2) + 3 = 3.
   13. (iii) For n = 3, t₃ = 3² − 2(3) + 3 = 6.
   14. (iii) For n = 4, t₄ = 4² − 2(4) + 3 = 11.
   15. (iii) For n = 5, t₅ = 5² − 2(5) + 3 = 18.

   Answer: (i) −1, 2, 5, 8, 11; (ii) −3, −8, −13, −18, −23; (iii) 2, 3, 6, 11, 18

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2. Ex 8.1 Q2

   10th and 15th terms of tₙ = 5n − 3 (n ≥ 1).

   1. The general term of the sequence is given by tₙ = 5n − 3.
   2. To find the 10th term, substitute n = 10 into the formula.
      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.

   3. t₁₀ = 5(10) − 3 = 50 − 3 = 47.
   4. To find the 15th term, substitute n = 15 into the formula.
      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. t₁₅ = 5(15) − 3 = 75 − 3 = 72.

   Answer: t₁₀ = 47; t₁₅ = 72

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3. Ex 8.1 Q3

   Are 97 and 172 terms of tₙ = 5n − 3 (n ≥ 1)?

   1. Set the general term equal to 97: 5n − 3 = 97.
   2. Add 3 to both sides: 5n = 100.
      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.

   3. Divide by 5: n = 20.
   4. Since n = 20 is a positive integer, 97 is the 20th term.
   5. Set the general term equal to 172: 5n − 3 = 172.
   6. Add 3 to both sides: 5n = 175.
      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. Divide by 5: n = 35.
   8. Since n = 35 is a positive integer, 172 is the 35th term.

   Answer: 97 is the 20th term; 172 is the 35th term

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4. Ex 8.1 Q4

   Which term of tₙ = 5n − 3 (n ≥ 1) is 607?

   1. The general term of the sequence is given as tₙ = 5n − 3.
   2. We are looking for the term that equals 607, so we set tₙ = 607.
   3. This gives the equation 5n − 3 = 607.
   4. Add 3 to both sides to get 5n = 610.
      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. Divide both sides by 5 to find n = 122.
      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. Since n = 122 is a positive integer, the 122nd term is 607.

   Answer: 122nd term

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5. Ex 8.1 Q5

   t₁ = −5, tₙ₊₁ = tₙ + 3 (n ≥ 1): first five terms; is 52 a term, and if so which one?

   1. The first term is given as t₁ = −5.
   2. The common difference is d = 3 because each term increases by 3.
      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.

   3. The second term is t₂ = −5 + 3 = −2.
   4. The third term is t₃ = −2 + 3 = 1.
   5. The fourth term is t₄ = 1 + 3 = 4.
   6. The fifth term is t₅ = 4 + 3 = 7.
   7. To check if 52 is a term, we use the formula for the nth term: tₙ = 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.

   8. Substituting the values, we get 52 = −5 + (n−1)3.
      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.

   9. Adding 5 to both sides gives 57 = (n−1)3.
      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.

   10. Dividing by 3 gives n−1 = 19.
   11. Adding 1 to both sides gives n = 20.
       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.

   12. Since n is a positive integer, 52 is the 20th term of the sequence.

   Answer: First five terms: −5, −2, 1, 4, 7; Yes, 52 is the 20th term

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6. Ex 8.1 Q6

   T₁ = 1, T₂ = 2, T₃ = 4 and Tₙ = Tₙ₋₁ + Tₙ₋₂ + Tₙ₋₃ for n ≥ 4. Find T₄ to T₈.

   1. Given T₁ = 1, T₂ = 2, T₃ = 4 and the recurrence relation Tₙ = Tₙ₋₁ + Tₙ₋₂ + Tₙ₋₃ for n ≥ 4.
   2. For n = 4, T₄ = T₃ + T₂ + T₁ = 4 + 2 + 1 = 7.
   3. For n = 5, T₅ = T₄ + T₃ + T₂ = 7 + 4 + 2 = 13.
   4. For n = 6, T₆ = T₅ + T₄ + T₃ = 13 + 7 + 4 = 24.
   5. For n = 7, T₇ = T₆ + T₅ + T₄ = 24 + 13 + 7 = 44.
   6. For n = 8, T₈ = T₇ + T₆ + T₅ = 44 + 24 + 13 = 81.

   Answer: T₄ = 7, T₅ = 13, T₆ = 24, T₇ = 44, T₈ = 81

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