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NCERT Class 10 Mathematics Exercise 5.2 Solutions: Arithmetic Progressions

AI-written, checked on 23 Oct 2026

Solutions written by Shishya, checked — not the board's official answers

Step-by-step working for every question of Exercise 5.2 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.

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How these were checked: the solutions were drafted with AI and then read and corrected on 23 Oct 2026. Every numerical final answer is recomputed from the question's own numbers by a calculator check that runs before each release; proofs and explanations were read, not machine-checked. The NCERT textbook is the authority — if something here disagrees with it, follow the book.

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Ex 5.2 Q1
Fill the missing value (a first term, d common difference, aₙ nth term): (i) a = 7, d = 3, n = 8, find aₙ (ii) a = −18, n = 10, aₙ = 0, find d (iii) d = −3, n = 18, aₙ = −5, find a (iv) a = −18.9, d = 2.5, aₙ = 3.6, find n (v) a = 3.5, d = 0, n = 105, find aₙ.
1. (i) 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.
2. (i) Substitute a = 7, d = 3, and n = 8 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. (i) Calculate a₈ = 7 + (8 − 1) × 3 = 7 + 21 = 28.
4. (ii) 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.
5. (ii) Substitute a = −18, n = 10, and aₙ = 0 into 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.
6. (ii) Solve 0 = −18 + (10 − 1)d for d.
7. (ii) Rearrange to get 9d = 18, so d = 2.
8. (iii) 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.
9. (iii) Substitute d = −3, n = 18, and aₙ = −5 into 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. (iii) Solve −5 = a + (18 − 1)(−3) for a.
11. (iii) Simplify to −5 = a − 51, so a = 46.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
12. (iv) 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.
13. (iv) Substitute a = −18.9, d = 2.5, and aₙ = 3.6 into 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.
14. (iv) Solve 3.6 = −18.9 + (n − 1) × 2.5 for n.
15. (iv) Rearrange to get 22.5 = (n − 1) × 2.5.
16. (iv) Divide by 2.5 to get n − 1 = 9, so n = 10.
17. (v) 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.
18. (v) Substitute a = 3.5, d = 0, and n = 105 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.
19. (v) Calculate a₁₀₅ = 3.5 + (105 − 1) × 0 = 3.5 + 0 = 3.5.
Answer: (i) 28; (ii) d = 2; (iii) a = 46; (iv) n = 10; (v) 3.5
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Ex 5.2 Q2
Choose and justify: (i) the 30th term of 10, 7, 4, … is (A) 97 (B) 77 (C) −77 (D) −87 (ii) the 11th term of −3, −1/2, 2, … is (A) 28 (B) 22 (C) −38 (D) −48 1/2.
1. (i) The first term a is 10 and the common difference d is 7 − 10 = −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. (i) The formula for the nth term 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. (i) Substituting n = 30, we get a₃₀ = 10 + (30 − 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. (i) Simplifying, a₃₀ = 10 + 29(−3) = 10 − 87 = −77.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
5. (i) This matches option (C).
6. (ii) The first term a is −3 and the common difference d is −1/2 − (−3) = 5/2.
  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.
7. (ii) Using the formula aₙ = a + (n − 1)d for n = 11.
  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. (ii) Substituting values, a₁₁ = −3 + (11 − 1)(5/2).
  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. (ii) Simplifying, a₁₁ = −3 + 10(5/2) = −3 + 25 = 22.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
10. (ii) This matches option (B).
Answer: (i) (C); (ii) (B)
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Ex 5.2 Q3
Fill the boxes in each AP: (i) 2, □, 26 (ii) □, 13, □, 3 (iii) 5, □, □, 9 1/2 (iv) −4, □, □, □, □, 6 (v) □, 38, □, □, □, −22.
1. (i) 2, □, 26: the middle term of three terms is the average of the other two: (2 + 26)/2 = 14.
2. (ii) □, 13, □, 3: a₂ = 13 and a₄ = 3 give 2d = 3 − 13 = −10, so d = −5. Then a₁ = 13 + 5 = 18 and a₃ = 13 − 5 = 8.
3. (iii) 5, □, □, 9 1/2: a₄ − a₁ = 3d, so 3d = 9.5 − 5 = 4.5 and d = 1.5. The missing terms are 5 + 1.5 = 6.5 = 6 1/2 and 6.5 + 1.5 = 8.
4. (iv) −4, □, □, □, □, 6: a₆ − a₁ = 5d, so 5d = 6 − (−4) = 10 and d = 2. The missing terms are −2, 0, 2, 4.
5. (v) □, 38, □, □, □, −22: a₆ − a₂ = 4d, so 4d = −22 − 38 = −60 and d = −15.
6. (v) a₁ = 38 + 15 = 53, a₃ = 38 − 15 = 23, a₄ = 23 − 15 = 8, a₅ = 8 − 15 = −7.
Answer: (i) 14; (ii) 18, 8; (iii) 6.5, 8 (that is, 6 1/2 and 8); (iv) −2, 0, 2, 4; (v) 53, 23, 8, −7
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Ex 5.2 Q4
Which term of 3, 8, 13, 18, … is 78?
1. The first term a is 3 and the common difference d is 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. Let the required term be the nth term, so aₙ = 78.
  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. Use the formula for the nth term of an AP: 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.
4. Substitute the values: 78 = 3 + (n − 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.
5. Subtract 3 from both sides: 75 = (n − 1) × 5.
  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. Divide both sides by 5: 15 = 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.
7. Add 1 to both sides: n = 16.
  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: 16th term
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Ex 5.2 Q5
Find the number of terms: (i) 7, 13, 19, …, 205 (ii) 18, 15 1/2, 13, …, −47.
1. (i) Identify the first term a = 7 and the common difference d = 13 − 7 = 6.
  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. (i) Use the formula for the nth term of an AP: 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. (i) Substitute the values: 205 = 7 + (n − 1)6.
  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. (i) Simplify: 198 = (n − 1)6, so n − 1 = 33, giving n = 34.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
5. (ii) Identify the first term a = 18 and the common difference d = 15.5 − 18 = −2.5.
  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.
6. (ii) 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.
7. (ii) Substitute the values: −47 = 18 + (n − 1)(−2.5).
  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.
8. (ii) Simplify: −65 = (n − 1)(−2.5), so n − 1 = 26, giving n = 27.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
Answer: (i) 34; (ii) 27
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Ex 5.2 Q6
Is −150 a term of the AP 11, 8, 5, 2, …?
1. Identify the first term a = 11 and the common difference 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.
2. Use the formula for the nth term of an AP: 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 given term −150 into the formula: −150 = 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.
4. Simplify the equation: −150 = 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.
5. Combine constants: −150 = 14 − 3n.
6. Rearrange to solve for n: 3n = 14 + 150.
7. Calculate the sum: 3n = 164.
8. Divide by 3: n = 164 ÷ 3.
9. Since 164 is not divisible by 3, n is not an integer.
10. The term number must be a positive integer, so −150 is not a term of the AP.
Answer: not a term
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Ex 5.2 Q7
An AP's 11th term is 38 and 16th term is 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 as 38, so a + 10d = 38.
3. The 16th term is given as 73, so a + 15d = 73.
4. Subtract the first equation from the second to find d: (a + 15d) - (a + 10d) = 73 - 38.
5. This simplifies to 5d = 35, so d = 7.
  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 = 7 into the first equation: a + 10(7) = 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 + 70 = 38, so a = -32.
8. The 31st term is given by the formula a + 30d.
9. Substitute the values: -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
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Ex 5.2 Q8
An AP has 50 terms; the 3rd term is 12 and the last is 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 50th term (last term) is 106, so a + 49d = 106.
4. Subtract the first equation from the second: (a + 49d) − (a + 2d) = 106 − 12.
5. This gives 47d = 94.
6. Solving for d, we get d = 94 ÷ 47 = 2.
7. Substitute d = 2 into a + 2d = 12 to find a.
  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. a + 2(2) = 12, so a + 4 = 12.
9. Thus, a = 8.
10. The 29th term is given by a + 28d.
11. Substitute the values: 8 + 28(2).
  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. Calculate the result: 8 + 56 = 64.
Answer: 64
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Ex 5.2 Q9
The 3rd and 9th terms of an AP are 4 and −8. Which term is zero?
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 a + 2d = 4.
3. The 9th term is given as a + 8d = -8.
4. Subtract the first equation from the second to find 6d = -12.
5. Solving for d gives d = -2.
6. Substitute d = -2 into a + 2d = 4 to get a - 4 = 4.
  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 = 8.
8. To find which term is zero, set the general term a + (n-1)d = 0.
  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.
9. Substitute the values to get 8 + (n-1)(-2) = 0.
  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. Simplify the equation to 8 - 2n + 2 = 0, which becomes 10 - 2n = 0.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
11. Solving for n gives 2n = 10, so n = 5.
Answer: 5th term
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Ex 5.2 Q10
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
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 17th term is given by a + 16d.
3. The 10th term is given by a + 9d.
4. According to the question, the 17th term exceeds the 10th term by 7.
5. So, (a + 16d) − (a + 9d) = 7.
6. Simplifying the left side gives 7d = 7.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
7. Dividing both sides by 7 gives d = 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.
Answer: d = 1
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Ex 5.2 Q11
Which term of 3, 15, 27, 39, … is 132 more than its 54th term?
1. The given arithmetic progression is 3, 15, 27, 39, ...
2. First term a = 3 and common difference d = 15 − 3 = 12
  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 54th term is a₅₄ = a + (54 − 1)d = 3 + 53 × 12 = 3 + 636 = 639
4. We need the term that is 132 more than the 54th term, so its value is 639 + 132 = 771
5. Let this be the nth term: aₙ = a + (n − 1)d = 771
  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.
6. Substitute values: 3 + (n − 1) × 12 = 771
  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. (n − 1) × 12 = 771 − 3 = 768
8. n − 1 = 768 ÷ 12 = 64
9. n = 64 + 1 = 65
Answer: 65th term
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Ex 5.2 Q12
Two APs have the same common difference and their 100th terms differ by 100. By how much do their 1000th terms differ?
1. Let the first AP have first term a and common difference d, and the second AP have first term b and common difference 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 100th term of the first AP is a + 99d and the 100th term of the second AP is b + 99d.
3. The difference between their 100th terms is (a + 99d) - (b + 99d) = a - b.
4. We are given that this difference is 100, so a - b = 100.
5. The 1000th term of the first AP is a + 999d and the 1000th term of the second AP is b + 999d.
6. The difference between their 1000th terms is (a + 999d) - (b + 999d) = a - b.
7. Since a - b = 100, the difference between the 1000th terms is also 100.
Answer: 100
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Ex 5.2 Q13
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 and common difference d = 7.
  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, a_n = a + (n − 1)d, we have 994 = 105 + (n − 1) × 7.
  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 105 from both sides gives 889 = (n − 1) × 7.
  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 both sides by 7 gives 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.
7. Adding 1 to both sides gives 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
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Ex 5.2 Q14
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. The last term is l = 248.
5. Using the formula for the nth term of an AP, 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.
6. Substituting the values, 248 = 12 + (n − 1)4.
  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. Simplifying, 248 − 12 = 4(n − 1).
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
8. 236 = 4(n − 1).
9. Dividing both sides by 4, 59 = 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.
10. Therefore, n = 60.
Answer: 60
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Ex 5.2 Q15
For which n are the nth terms of 63, 65, 67, … and 3, 10, 17, … equal?
1. For the first AP, the first term a is 63 and the common difference d is 65 - 63 = 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.
2. The formula for the nth term of the first AP is a_n = 63 + (n - 1) × 2.
  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. Simplifying this gives a_n = 63 + 2n - 2 = 2n + 61.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
4. For the second AP, the first term a is 3 and the common difference d is 10 - 3 = 7.
  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. The formula for the nth term of the second AP is b_n = 3 + (n - 1) × 7.
  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.
6. Simplifying this gives b_n = 3 + 7n - 7 = 7n - 4.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
7. We equate the two nth terms to find when they are equal: 2n + 61 = 7n - 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.
8. Rearranging the terms gives 61 + 4 = 7n - 2n.
9. This simplifies to 65 = 5n.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
10. Solving for n gives n = 65 ÷ 5 = 13.
Answer: n = 13
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Ex 5.2 Q16
Find the AP whose 3rd term is 16 and whose 7th term exceeds the 5th by 12.
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 exceeds the 5th term by 12, so (a + 6d) - (a + 4d) = 12.
4. Simplifying the second equation gives 2d = 12, which means 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 the first equation: a + 2(6) = 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 + 12 = 16, so a = 4.
7. The AP is formed by starting with 4 and adding 6 repeatedly: 4, 10, 16, 22, ...
Answer: 4, 10, 16, 22
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Ex 5.2 Q17
Find the 20th term from the end of the AP 3, 8, 13, …, 253.
1. The given AP is 3, 8, 13, …, 253.
2. Here, first term a = 3 and 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.
3. Let the total number of terms be n.
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, 253 = 3 + (n − 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. Simplifying, 250 = 5(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. Dividing by 5, 50 = n − 1.
8. So, n = 51.
9. The 20th term from the end is the (n − 20 + 1)th term from the beginning.
10. This is the (51 − 20 + 1)th term, which is the 32nd term.
11. The 32nd term is a + (32 − 1)d.
12. Substituting values, 3 + 31 × 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.
13. Calculating, 3 + 155 = 158.
Answer: 158
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Ex 5.2 Q18
The 4th and 8th terms of an AP add to 24 and the 6th and 10th add to 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.
3. Their sum is (a + 3d) + (a + 7d) = 2a + 10d = 24.
4. Simplifying gives a + 5d = 12.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
5. The 6th term is a + 5d and the 10th term is a + 9d.
6. Their sum is (a + 5d) + (a + 9d) = 2a + 14d = 44.
7. Simplifying gives a + 7d = 22.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
8. Subtracting the equation a + 5d = 12 from a + 7d = 22 gives 2d = 10.
9. Solving for d gives d = 5.
10. Substituting d = 5 into a + 5d = 12 gives a + 25 = 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.
11. Solving for a gives a = -13.
12. The first term is -13, the second term is -13 + 5 = -8, and the third term is -8 + 5 = -3.
Answer: -13, -8, -3
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Ex 5.2 Q19
Subba Rao started in 1995 on ₹5000 a year with a ₹200 rise each year. In which year did his income reach ₹7000?
1. The initial income in 1995 is ₹5000, so the first term a = 5000.
2. The annual increase is ₹200, so the common difference d = 200.
  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. We need to find the year when the income reaches ₹7000, so the nth term aₙ = 7000.
  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. Using the formula for the nth term of an AP, 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.
5. Substituting the values, 7000 = 5000 + (n − 1)200.
  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. Subtracting 5000 from both sides gives 2000 = (n − 1)200.
  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. Dividing both sides by 200 gives 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. Adding 1 to both sides gives n = 11.
  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. This means the income reaches ₹7000 in the 11th year.
10. Counting 11 years starting from 1995 (where 1995 is the 1st year), the 11th year is 1995 + 10 = 2005.
Answer: 2005
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Ex 5.2 Q20
Ramkali saved ₹5 in week 1 and ₹1.75 more each week. In which week n did she save ₹20.75?
1. The first term a is ₹5 and the common difference d is ₹1.75.
  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. We need to find the week number n where the amount saved is ₹20.75, so the nth term aₙ is 20.75.
  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. 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.
4. Substitute the known values into the formula: 20.75 = 5 + (n − 1) × 1.75.
  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. Subtract 5 from both sides to isolate the term with n: 20.75 − 5 = (n − 1) × 1.75.
  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. Calculate the difference: 15.75 = (n − 1) × 1.75.
7. Divide both sides by 1.75 to solve for (n − 1): n − 1 = 15.75 ÷ 1.75.
  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. Perform the division: n − 1 = 9.
9. Add 1 to both sides to find n: n = 9 + 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.
10. The value of n is 10.
Answer: n = 10
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Other exercises in this chapter

Exercise 5.1 — 4 questions
Exercise 5.3 — 20 questions
Exercise 5.4 — 5 questions

← Arithmetic Progressions: notes, practice and all exercises

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NCERT Class 10 Mathematics Exercise 5.2 Solutions: Arithmetic Progressions

Ex 5.2 Q1

Ex 5.2 Q2

Ex 5.2 Q3

Ex 5.2 Q4

Ex 5.2 Q5

Ex 5.2 Q6

Ex 5.2 Q7

Ex 5.2 Q8

Ex 5.2 Q9

Ex 5.2 Q10

Ex 5.2 Q11

Ex 5.2 Q12

Ex 5.2 Q13

Ex 5.2 Q14

Ex 5.2 Q15

Ex 5.2 Q16

Ex 5.2 Q17

Ex 5.2 Q18

Ex 5.2 Q19

Ex 5.2 Q20