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NCERT Class 10 Mathematics Exercise 5.4 Solutions: Arithmetic Progressions

Step-by-step working for every question of Exercise 5.4 in NCERT's Class 10 Mathematics book, chapter “Arithmetic Progressions”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Ex 5.4 Q1

   Which term of 121, 117, 113, … is its first negative term?

   1. The first term a is 121 and the common difference d is 117 - 121 = -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.

   2. The formula for the nth term of an arithmetic progression is 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 to get aₙ = 121 + (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.

   4. Simplify the expression to find aₙ = 121 - 4n + 4 = 125 - 4n.
      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. For the term to be negative, we need aₙ < 0, so 125 - 4n < 0.
   6. Rearrange the inequality to get 4n > 125.
   7. Divide by 4 to find n > 31.25.
   8. Since n must be a whole number, the smallest integer greater than 31.25 is 32.

   Answer: 32nd term

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2. Ex 5.4 Q2

   The 3rd and 7th terms of an AP add to 6 and multiply to 8. Find the sum of the first 16 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 3rd term is a + 2d and the 7th term is a + 6d.
   3. Given their sum is 6, so (a + 2d) + (a + 6d) = 6.
   4. Simplifying gives 2a + 8d = 6, which means a + 4d = 3.
      why?

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

   5. Given their product is 8, so (a + 2d)(a + 6d) = 8.
   6. Expanding gives a² + 8ad + 12d² = 8.
      why?

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

   7. Substitute a = 3 - 4d into the product equation.
      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. (3 - 4d + 2d)(3 - 4d + 6d) = 8.
   9. (3 - 2d)(3 + 2d) = 8.
   10. 9 - 4d² = 8.
   11. 4d² = 1, so d² = 1/4.
   12. Thus d = 1/2 or d = -1/2.
   13. Case 1: If d = 1/2, then a = 3 - 4(1/2) = 1.
   14. Sum of first 16 terms S₁₆ = 16/2 [2(1) + 15(1/2)].
       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.

   15. S₁₆ = 8 [2 + 7.5] = 8 × 9.5 = 76.
       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.

   16. Case 2: If d = -1/2, then a = 3 - 4(-1/2) = 5.
   17. Sum of first 16 terms S₁₆ = 16/2 [2(5) + 15(-1/2)].
       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.

   18. S₁₆ = 8 [10 - 7.5] = 8 × 2.5 = 20.
       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.

   Answer: 76 (when d = 1/2), 20 (when d = -1/2)

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3. Ex 5.4 Q3

   A ladder's rungs are 25 cm apart and shrink evenly from 45 cm at the bottom to 25 cm at the top; the top and bottom rungs are 2 1/2 m apart. Find the total length of wood for the rungs.

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

   1. The distance between the bottom and top rungs is 2.5 m, which equals 250 cm.
   2. Since the rungs are spaced 25 cm apart, the number of gaps is 250 ÷ 25 = 10.
   3. The total number of rungs is the number of gaps plus one, so n = 10 + 1 = 11.
   4. The length of the bottom rung is a = 45 cm and the top rung is l = 25 cm.
   5. The lengths form an arithmetic progression, so the total length of wood is the sum of these 11 terms.
   6. Using the sum formula S = n/2 × (a + l), we get S = 11/2 × (45 + 25).
      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. Simplifying the expression gives S = 11/2 × 70.
      why?

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

   8. Calculating the final value gives S = 11 × 35 = 385 cm.

   Answer: 385 cm

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4. Ex 5.4 Q4

   Houses are numbered 1 to 49 in a row. Show there is an x for which the house numbers before x add up to the house numbers after x, and find x.

   1. The total sum of house numbers from 1 to 49 is S = 49 × 50 ÷ 2 = 1225.
   2. Let the required house number be x.
   3. The sum of numbers before x is 1 + 2 + ... + (x − 1) = x(x − 1) ÷ 2.
   4. The sum of numbers after x is (x + 1) + ... + 49 = 1225 − x(x + 1) ÷ 2.
   5. Equating the two sums gives x(x − 1) ÷ 2 = 1225 − x(x + 1) ÷ 2.
   6. Multiplying by 2 yields x(x − 1) = 2450 − x(x + 1).
   7. Expanding and simplifying gives 2x² = 2450, so x² = 1225.
      why?

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

   8. Taking the positive square root gives x = 35.

   Answer: x = 35

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5. Ex 5.4 Q5

   A terrace has 15 concrete steps, each 50 m long, with a rise of 1/4 m and a tread of 1/2 m. Find the total volume of concrete.

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

   1. The terrace has 15 steps, each 50 m long.
   2. The rise of each step is 1/4 m and the tread is 1/2 m.
   3. The volume of the first step is 50 × 1/4 × 1/2 = 6.25 m³.
   4. The volume of the second step is 50 × (2/4) × 1/2 = 12.5 m³.
   5. The volume of the third step is 50 × (3/4) × 1/2 = 18.75 m³.
   6. The volumes of the 15 steps form an arithmetic progression with first term 6.25 and common difference 6.25.
      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.

   7. The sum of the volumes is given by 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.

   8. Substituting the values, S = 15/2 × [2 × 6.25 + 14 × 6.25].
      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. Simplifying the expression inside the brackets gives 12.5 + 87.5 = 100.
      why?

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

   10. The total volume is 15/2 × 100 = 750 m³.

   Answer: 750 m³

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