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NCERT Class 10 Mathematics Exercise 5.1 Solutions: Arithmetic Progressions

Step-by-step working for every question of Exercise 5.1 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 5.1 Q1

   Do these lists form an AP? (i) taxi fare after each km: ₹15 for the first km, ₹8 for each further km (ii) air left in a cylinder when a pump removes 1/4 of the remaining air each time (iii) cost of digging a well after each metre: ₹150 for the first metre, rising by ₹50 per metre (iv) money in an account each year when ₹10000 earns 8% compound interest.

   1. (i) The first term is 15 and the common difference is 8, so the list is 15, 23, 31, ... which forms an AP.
      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. (ii) The air left after the first removal is 3/4 of the total, and after the second is 9/16 of the total, so the differences are not constant and it is not an AP.
   3. (iii) The first term is 150 and the common difference is 50, so the list is 150, 200, 250, ... which forms an AP.
      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. (iv) The amount grows by 8% of the previous year's balance, so the differences increase each year and it is not an AP.
      why?

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

   Answer: (i) an AP; (ii) not an AP; (iii) an AP; (iv) not an AP

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2. Ex 5.1 Q2

   Write the first four terms of the AP: (i) a = 10, d = 10 (ii) a = −2, d = 0 (iii) a = 4, d = −3 (iv) a = −1, d = 1/2 (v) a = −1.25, d = −0.25.

   1. (i) The first term is a = 10.
   2. (i) The second term is a + d = 10 + 10 = 20.
   3. (i) The third term is a + 2d = 10 + 20 = 30.
   4. (i) The fourth term is a + 3d = 10 + 30 = 40.
   5. (ii) The first term is a = −2.
   6. (ii) The second term is a + d = −2 + 0 = −2.
   7. (ii) The third term is a + 2d = −2 + 0 = −2.
   8. (ii) The fourth term is a + 3d = −2 + 0 = −2.
   9. (iii) The first term is a = 4.
   10. (iii) The second term is a + d = 4 + (−3) = 1.
   11. (iii) The third term is a + 2d = 4 + (−6) = −2.
   12. (iii) The fourth term is a + 3d = 4 + (−9) = −5.
   13. (iv) The first term is a = −1.
   14. (iv) The second term is a + d = −1 + 1/2 = −1/2.
   15. (iv) The third term is a + 2d = −1 + 1 = 0.
   16. (iv) The fourth term is a + 3d = −1 + 3/2 = 1/2.
   17. (v) The first term is a = −1.25.
   18. (v) The second term is a + d = −1.25 + (−0.25) = −1.5.
   19. (v) The third term is a + 2d = −1.25 + (−0.5) = −1.75.
   20. (v) The fourth term is a + 3d = −1.25 + (−0.75) = −2.

   Answer: (i) 10, 20, 30, 40; (ii) -2, -2, -2, -2; (iii) 4, 1, -2, -5; (iv) -1, -1/2, 0, 1/2; (v) -1.25, -1.5, -1.75, -2

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3. Ex 5.1 Q3

   Write the first term and common difference: (i) 3, 1, −1, −3, … (ii) −5, −1, 3, 7, … (iii) 1/3, 5/3, 9/3, 13/3, … (iv) 0.6, 1.7, 2.8, 3.9, …

   1. (i) The first term a is the first number in the sequence, which is 3.
   2. (i) The common difference d is found by subtracting the first term from the second term: 1 − 3 = −2.
      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. (ii) The first term a is the first number in the sequence, which is −5.
   4. (ii) The common difference d is found by subtracting the first term from the second term: −1 − (−5) = −1 + 5 = 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.

   5. (iii) The first term a is the first number in the sequence, which is 1/3.
   6. (iii) The common difference d is found by subtracting the first term from the second term: 5/3 − 1/3 = 4/3.
      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.

   7. (iv) The first term a is the first number in the sequence, which is 0.6.
   8. (iv) The common difference d is found by subtracting the first term from the second term: 1.7 − 0.6 = 1.1.
      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.

   Answer: (i) a = 3, d = -2; (ii) a = -5, d = 4; (iii) a = 1/3, d = 4/3; (iv) a = 0.6, d = 1.1

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4. Ex 5.1 Q4

   Which are APs? For an AP give d and three more terms: (i) 2, 4, 8, 16 (ii) 2, 5/2, 3, 7/2 (iii) −1.2, −3.2, −5.2, −7.2 (iv) −10, −6, −2, 2 (v) 3, 3 + √2, 3 + 2√2, 3 + 3√2 (vi) 0.2, 0.22, 0.222, 0.2222 (vii) 0, −4, −8, −12 (viii) −1/2, −1/2, −1/2, −1/2 (ix) 1, 3, 9, 27 (x) a, 2a, 3a, 4a (xi) a, a², a³, a⁴ (xii) √2, √8, √18, √32 (xiii) √3, √6, √9, √12 (xiv) 1², 3², 5², 7² (xv) 1², 5², 7², 73.

   1. (i) 2, 4, 8, 16: Differences are 4−2=2, 8−4=4, 16−8=8. Since differences are not equal, it is not an AP.
   2. (ii) 2, 5/2, 3, 7/2: Differences are 5/2−2=1/2, 3−5/2=1/2, 7/2−3=1/2. It is an AP with d = 1/2. Next terms: 7/2+1/2=4, 4+1/2=9/2, 9/2+1/2=5.
   3. (iii) −1.2, −3.2, −5.2, −7.2: Differences are −3.2−(−1.2)=−2, −5.2−(−3.2)=−2, −7.2−(−5.2)=−2. It is an AP with d = −2. Next terms: −7.2−2=−9.2, −9.2−2=−11.2, −11.2−2=−13.2.
   4. (iv) −10, −6, −2, 2: Differences are −6−(−10)=4, −2−(−6)=4, 2−(−2)=4. It is an AP with d = 4. Next terms: 2+4=6, 6+4=10, 10+4=14.
   5. (v) 3, 3+√2, 3+2√2, 3+3√2: Differences are (3+√2)−3=√2, (3+2√2)−(3+√2)=√2, (3+3√2)−(3+2√2)=√2. It is an AP with d = √2. Next terms: 3+3√2+√2=3+4√2, 3+4√2+√2=3+5√2, 3+5√2+√2=3+6√2.
   6. (vi) 0.2, 0.22, 0.222, 0.2222: Differences are 0.22−0.2=0.02, 0.222−0.22=0.002, 0.2222−0.222=0.0002. Since differences are not equal, it is not an AP.
   7. (vii) 0, −4, −8, −12: Differences are −4−0=−4, −8−(−4)=−4, −12−(−8)=−4. It is an AP with d = −4. Next terms: −12−4=−16, −16−4=−20, −20−4=−24.
   8. (viii) −1/2, −1/2, −1/2, −1/2: Differences are −1/2−(−1/2)=0, −1/2−(−1/2)=0, −1/2−(−1/2)=0. It is an AP with d = 0. Next terms: −1/2, −1/2, −1/2.
   9. (ix) 1, 3, 9, 27: Differences are 3−1=2, 9−3=6, 27−9=18. Since differences are not equal, it is not an AP.
   10. (x) a, 2a, 3a, 4a: Differences are 2a−a=a, 3a−2a=a, 4a−3a=a. It is an AP with d = a. Next terms: 4a+a=5a, 5a+a=6a, 6a+a=7a.
   11. (xi) a, a², a³, a⁴: Differences are a²−a, a³−a², a⁴−a³. These are not equal in general, so it is not an AP.
   12. (xii) √2, √8, √18, √32: Simplify terms: √2, 2√2, 3√2, 4√2. Differences are 2√2−√2=√2, 3√2−2√2=√2, 4√2−3√2=√2. It is an AP with d = √2. Next terms: 4√2+√2=5√2, 5√2+√2=6√2, 6√2+√2=7√2.
       why?

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

   13. (xiii) √3, √6, √9, √12: Differences are √6−√3, √9−√6, √12−√9. These are not equal, so it is not an AP.
   14. (xiv) 1², 3², 5², 7²: Terms are 1, 9, 25, 49. Differences are 9−1=8, 25−9=16, 49−25=24. Since differences are not equal, it is not an AP.
   15. (xv) 1², 5², 7², 73: Terms are 1, 25, 49, 73. Differences are 25−1=24, 49−25=24, 73−49=24. It is an AP with d = 24. Next terms: 73+24=97, 97+24=121, 121+24=145.

   Answer: (i) not an AP; (ii) d = 1/2; 4, 9/2, 5; (iii) d = -2; -9.2, -11.2, -13.2; (iv) d = 4; 6, 10, 14; (v) d = √2; 3 + 4√2, 3 + 5√2, 3 + 6√2; (vi) not an AP; (vii) d = -4; -16, -20, -24; (viii) d = 0; -1/2, -1/2, -1/2; (ix) not an AP; (x) d = a; 5a, 6a, 7a; (xi) not an AP; (xii) d = √2; 5√2, 6√2, 7√2; (xiii) not an AP; (xiv) not an AP; (xv) d = 24; 97, 121, 145

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