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NCERT Class 9 Mathematics Exercise Set 8.2 Solutions: Predicting What Comes Next: Exploring Sequences and Progressions

Step-by-step working for every question of Exercise Set 8.2 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 8.2 Q1

   10th and 26th terms of the AP 3, 8, 13, 18, …

   1. Identify the first term a = 3 and the common difference d = 8 − 3 = 5.
      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. Use the formula for the nth term of an AP: 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.

   3. For the 10th term, substitute n = 10: t₁₀ = 3 + (10 − 1) × 5.
      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 t₁₀ = 3 + 9 × 5 = 3 + 45 = 48.
   5. For the 26th term, substitute n = 26: t₂₆ = 3 + (26 − 1) × 5.
      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 t₂₆ = 3 + 25 × 5 = 3 + 125 = 128.

   Answer: t₁₀ = 48; t₂₆ = 128

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2. Ex 8.2 Q2

   AP 21, 18, 15, …: which term is −81? Is 0 a term? Why?

   1. The first term a is 21 and the common difference d is 18 − 21 = −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.

   2. To find the term equal to −81, use the formula aₙ = 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.

   3. Substitute the values: −81 = 21 + (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.

   4. Simplify the equation: −81 = 21 − 3n + 3.
      why?

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

   5. Combine constants: −81 = 24 − 3n.
   6. Rearrange to solve for n: 3n = 24 + 81.
   7. Calculate n: 3n = 105, so n = 35.
   8. Since n is a positive integer, −81 is the 35th term.
   9. To check if 0 is a term, set aₙ = 0 in the formula.
      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.

   10. Substitute values: 0 = 21 + (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.

   11. Simplify: 0 = 21 − 3n + 3.
       why?

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

   12. Combine constants: 0 = 24 − 3n.
   13. Rearrange: 3n = 24.
   14. Calculate n: n = 8.
   15. Since n is a positive integer, 0 is the 8th term.

   Answer: −81 is the 35th term; 0 is the 8th term

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3. Ex 8.2 Q3

   nth term and a recursive rule of the AP 11, 8, 5, 2, …

   1. Identify the first term of the arithmetic progression as a = 11.
   2. Find the common difference d by subtracting the first term from the second term: d = 8 − 11 = −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. Use the formula for the nth term of an AP, 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.

   4. Substitute the values a = 11 and d = −3 into the formula: tₙ = 11 + (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.

   5. Simplify the expression: tₙ = 11 − 3n + 3.
      why?

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

   6. Combine like terms to get the final formula: tₙ = 14 − 3n.
   7. Verify the first term by substituting n = 1: t₁ = 14 − 3(1) = 11.
      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.

   8. Determine the recursive rule by observing that each term is 3 less than the previous term.
   9. Write the recursive rule as tₙ₊₁ = tₙ − 3.

   Answer: tₙ = 14 − 3n; t₁ = 11; tₙ₊₁ = tₙ − 3

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4. Ex 8.2 Q4

   An AP has 50 terms, 3rd term 12 and last term 106. Find the 29th 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 12, so a + 2d = 12.
   3. The last term (50th term) is 106, so a + 49d = 106.
   4. Subtract the first equation from the second: (a + 49d) - (a + 2d) = 106 - 12.
   5. This simplifies to 47d = 94, so d = 2.
      why?

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

   6. Substitute d = 2 into a + 2d = 12 to get a + 4 = 12, so a = 8.
      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. The 29th term is given by a + 28d.
   8. Substitute the values: 8 + 28(2) = 8 + 56 = 64.
      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.

   Answer: a = 8, d = 2, t₂₉ = 64

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5. Ex 8.2 Q5

   How many 2-digit numbers are divisible by 3, and what is their sum?

   1. The 2-digit numbers divisible by 3 form an arithmetic progression starting from 12 and ending at 99.
   2. Here, the first term a = 12, the common difference d = 3, and the last term l = 99.
      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. Using the formula for the nth term, l = a + (n − 1)d, we substitute the values to get 99 = 12 + (n − 1)3.
      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.

   4. Solving for n, we get 87 = 3(n − 1), which simplifies to n − 1 = 29, so n = 30.
      why?

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

   5. Thus, there are 30 such numbers.
   6. The sum of these numbers is given by S = n/2 × (a + l).
      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.

   7. Substituting the values, S = 30/2 × (12 + 99).
      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.

   8. This simplifies to S = 15 × 111.
      why?

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

   9. Calculating the product gives S = 1665.

   Answer: There are 30 such numbers and their sum is 1665.

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6. Ex 8.2 Q6

   Salary starts at ₹5,00,000 a year with an increment of ₹20,000 every year. After how many years does it reach ₹7,00,000?

   1. The initial salary is the first term, so a = 500000.
   2. The annual increment is the common difference, so d = 20000.
      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 target salary is the nth term, so a_n = 700000.
      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.

   4. Use the formula for the nth term of an arithmetic progression: a_n = 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. Substitute the values: 700000 = 500000 + (n − 1) × 20000.
      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 the equation: 200000 = (n − 1) × 20000.
      why?

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

   7. Divide both sides by 20000: 10 = 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.

   8. Solve for n: n = 11.
   9. This means the salary reaches 7,00,000 in the 11th year.
   10. The number of years after the start is 11 − 1 = 10 years.

   Answer: 10 years; 11th year

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7. Ex 8.2 Q7

   Marbles in rows of 1, 2, 3, … up to 25 rows. Total number of marbles?

   1. The number of marbles in the rows forms an arithmetic progression with first term a = 1 and common difference d = 1.
      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 total number of rows is n = 25.
   3. The sum of the first n terms of an AP is given by the formula S = n/2 × [2a + (n − 1)d].
      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.

   4. Substitute the values into the formula: S = 25/2 × [2(1) + (25 − 1)(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.

   5. Simplify the expression inside the brackets: 2 + 24 = 26.
      why?

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

   6. Calculate the final sum: S = 25/2 × 26 = 25 × 13 = 325.

   Answer: 325

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