--- title: "Arithmetic Progressions: CBSE Class 10 Maths previous year questions" url: https://www.swavid.com/cbse/class-10/maths/pyq/arithmetic-progressions --- # Arithmetic Progressions: CBSE Class 10 Maths previous year questions 28 questions from CBSE Class 10 board papers, newest first, each with full working. ## CBSE Class 10 Maths Standard Question Paper 2026 (Set 30/1/1) with Solutions ### Question 6 *1 mark · MCQ* The common difference of the AP : $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots$ is : - \sqrt{2} - 1 - 2\sqrt{2} - -\sqrt{2} **Solution** 1. The given AP is $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots$. 2. The common difference $d$ is obtained by subtracting the first term from the second term: $d = 2\sqrt{2} - \sqrt{2} = \sqrt{2}$. **Answer:** (a) $\sqrt{2}$ > Common mistake: Subtracting terms incorrectly or getting confused with surds. ### Question 36 *4 marks · Case-based* In a potato race, a bucket is placed at the starting point, which is $5\text{ m}$ from the first potato. The other potatoes are arranged $3\text{ m}$ apart in a straight line, with a total of 10 potatoes, as shown in the figure : A competitor starts from the bucket, picks up the nearest potato, runs back to the bucket to drop it in, then returns to pick up the next potato. This process continues until all the potatoes are in the bucket. Based on the above information, answer the following questions : (i) What is the distance covered to pick up the first potato and drop it in bucket ? [1] (ii) What is the distance covered to pick up the second potato and drop it in bucket ? [1] (iii) (a) What is the total distance the competitor has to run ? [2] OR (iii) (b) If average speed of competitor is $5\text{ m/s}$, then find the average time taken by competitor to put all the potatoes in the bucket. [2] **Part (i)** 1. The distance of the first potato from the bucket is $5\text{ m}$. 2. The competitor runs to the potato and back to the bucket, so the total distance is $2 \times 5 = 10\text{ m}$. Answer (i): $10\text{ m}$ **Part (ii)** 1. The distance of the second potato from the bucket is $5 + 3 = 8\text{ m}$. 2. The total distance covered to pick up the second potato and drop it in the bucket is $2 \times 8 = 16\text{ m}$. Answer (ii): $16\text{ m}$ **Part (iii) (a)** 1. The distances form an arithmetic progression: $10, 16, 22, \dots$ with first term $a = 10$ and common difference $d = 6$. 2. The total distance for 10 potatoes is given by the sum of 10 terms of the AP: $S_{10} = \frac{10}{2} [2(10) + (10 - 1)6]$. 3. $S_{10} = 5 [20 + 54] = 5 \times 74 = 370\text{ m}$. Answer (iii) (a): $370\text{ m}$ **Part (iii) (b)** 1. The total distance covered is $370\text{ m}$ and the average speed is $5\text{ m/s}$. 2. Time taken = $\frac{\text{Total Distance}}{\text{Speed}} = \frac{370}{5} = 74\text{ s}$. Answer (iii) (b): $74\text{ seconds}$ **Answer:** Refer to individual parts for solutions. > Common mistake: Forgetting to multiply the distance to the potato by 2 for the round trip. ## CBSE Class 10 Maths Basic Question Paper 2025 (Set 430/1/1) with Solutions ### Question 37 (i) *1 mark · Case-based* What is the radius of the 13th spiral? **Part (i)** 1. The radii of the successive semicircles form an Arithmetic Progression: $0.5\text{ cm}, 1.0\text{ cm}, 1.5\text{ cm}, \dots$ 2. Here, the first term $a = 0.5$ and the common difference $d = 0.5$. 3. Using the nth term formula $a_n = a + (n - 1)d$, find the 13th term: $a_{13} = 0.5 + (13 - 1) \times 0.5 = 0.5 + 12 \times 0.5 = 0.5 + 6 = 6.5\text{ cm}$. Answer (i): $6.5\text{ cm}$ **Answer:** The radius of the 13th spiral is $6.5\text{ cm}$. > Common mistake: Using the lengths of the semicircles instead of their radii. ### Question 37 (ii) *1 mark · Case-based* If the radius of the nth spiral is 500 cm, find the value of n. **Part (ii)** 1. The radii of the successive semi-circles form an AP: $0.5, 1.0, 1.5, \dots$ with first term $a = 0.5$ and common difference $d = 0.5$. 2. The formula for the nth term of an AP is $a_n = a + (n - 1)d$. 3. Substitute the given radius $500\text{ cm}$ for $a_n$ and the values of $a$ and $d$: $500 = 0.5 + (n - 1)0.5$. 4. Solve for $n$: $499.5 = (n - 1)0.5 \implies n - 1 = 999 \implies n = 100$. Answer (ii): $n = 100$ **Answer:** $n = 100$ > Common mistake: Confusing the radius of the spiral with the total length of the spiral. ### Question 37 (iii) (a) *2 marks · Case-based* Find the total number of saplings till the 11th spiral. **Part (a)** 1. The lengths of the semi-circles form an AP with first term $a = 0.5\pi$ and common difference $d = 0.5\pi$. 2. The sum of the lengths of 11 spirals is given by $S_{11} = \frac{11}{2} [2a + (11-1)d]$. 3. Substituting the values, $S_{11} = \frac{11}{2} [2(0.5\pi) + 10(0.5\pi)] = \frac{11}{2} [6\pi] = 33\pi\text{ cm}$. 4. Using $\pi = \frac{22}{7}$, the total length is $33 \times \frac{22}{7} = \frac{726}{7}\text{ cm}$ or $143\pi\text{ cm}$. Answer (a): $143\pi\text{ cm}$ **Answer:** The total length of wire for 11 spirals is $143\pi\text{ cm}$. > Common mistake: Taking the diameter instead of the radius in the formula for the length of a semi-circle. ### Question 37 (OR) *2 marks · Case-based* Till which spiral, will there be a total of 450 saplings? **Solution** 1. Let the total number of saplings till the $n$-th spiral be $450$. 2. The lengths of the semicircles forming the spirals are given in AP with first term $a = 5$ and common difference $d = 5$. 3. Using the sum formula $S_n = \frac{n}{2} [2a + (n-1)d]$, we get $450 = \frac{n}{2} [2(5) + (n-1)5]$. 4. Simplifying the equation gives $900 = n[10 + 5n - 5] = n[5n + 5] = 5n(n+1)$. 5. Dividing by $5$, we get $n(n+1) = 180$, which means $n^2 + n - 180 = 0$. 6. Solving by factorisation: $n^2 + 15n - 12n - 180 = 0 \implies n(n+15) - 12(n+15) = 0$. 7. Thus, $(n-12)(n+15) = 0$, giving $n = 12$ (since $n$ cannot be negative). **Answer:** 12th spiral > Common mistake: Not rejecting the negative value of $n$. ## CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1) with Solutions ### Question 11 *1 mark · MCQ* If the sum of first $m$ terms of an AP is $2m^2 + 3m$, then its second term is : - 10 - 9 - 12 - 4 **Solution** 1. Given $S_m = 2m^2 + 3m$. The first term $a = S_1 = 2(1)^2 + 3(1) = 5$. 2. The sum of first two terms is $S_2 = 2(2)^2 + 3(2) = 8 + 6 = 14$. The second term is $a_2 = S_2 - S_1 = 14 - 5 = 9$. **Answer:** (b) 9 > Common mistake: Directly substituting $m = 2$ into the formula to find the second term instead of finding $S_2 - S_1$. ### Question 36 *4 marks · Case-based* A school is organizing a charity run to raise funds for a local hospital. The run is planned as a series of rounds around a track, with each round being 300 metres. To make the event more challenging and engaging, the organizers decide to increase the distance of each subsequent round by 50 metres. For example, the second round will be 350 metres, the third round will be 400 metres and so on. The total number of rounds planned is 10. (i) Write the fourth, fifth and sixth term of the Arithmetic Progression so formed. (ii) Determine the distance of the 8th round. (iii) (a) Find the total distance run after completing all 10 rounds. **Part (i)** 1. The first term $a = 300$ and common difference $d = 50$. 2. The fourth term is $a_4 = a + 3d = 300 + 3(50) = 450$. 3. The fifth term is $a_5 = a + 4d = 300 + 4(50) = 500$. 4. The sixth term is $a_6 = a + 5d = 300 + 5(50) = 550$. Answer (i): 450 m, 500 m, 550 m **Part (ii)** 1. The distance of the 8th round is given by the term $a_8$. 2. Using the formula $a_n = a + (n - 1)d$, we get $a_8 = 300 + (8 - 1)50$. 3. Substitute the values: $a_8 = 300 + 7(50) = 650$. Answer (ii): 650 metres **Part (iii)(a)** 1. The total distance after 10 rounds is the sum of the first 10 terms of the AP, $S_{10}$. 2. Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$, substitute $n = 10$, $a = 300$, and $d = 50$. 3. $S_{10} = \frac{10}{2}[2(300) + (10 - 1)50]$. 4. $S_{10} = 5[600 + 450] = 5(1050) = 5250$. Answer (iii)(a): 5250 metres **Answer:** Total distance after 10 rounds is 5250 metres. > Common mistake: Confusing the nth term formula with the sum of n terms formula. ### Question 36 (OR) *4 marks · Case-based* If a runner completes only the first 6 rounds, what is the total distance run by the runner? **Part (i)** 1. The total distance run in 6 rounds is the sum of the first 6 terms of the AP, $S_6$. 2. Here, first term $a = 300$, common difference $d = 50$, and number of terms $n = 6$. 3. Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$, substitute the values. 4. $S_6 = \frac{6}{2}[2(300) + (6 - 1)50]$. 5. $S_6 = 3[600 + 5(50)] = 3[600 + 250] = 3(850) = 2550$. Answer (i): 2550 metres **Answer:** Total distance for the first 6 rounds is 2250 metres. > Common mistake: Using $n=7$ instead of $n=6$ for the number of rounds. ## CBSE Class 10 Maths Basic Question Paper 2024 (Set 430/1/3) with Solutions ### Question 2 *1 mark · MCQ* In an A.P., if $a = 8$ and $a_{10} = -19$, then value of $d$ is : - $3$ - $-\frac{11}{9}$ - $-\frac{27}{10}$ - $-3$ **Solution** 1. The formula for the $n$-th term of an A.P. is $a_n = a + (n - 1)d$. 2. Substitute $a = 8$ and $a_{10} = -19$ into the formula to get $-19 = 8 + (10 - 1)d$. 3. Solving for $d$, we get $9d = -19 - 8 = -27$, which gives $d = -3$. **Answer:** (d) $-3$ > Common mistake: Taking $n=10$ as $10d$ instead of $9d$. ### Question 33 *5 marks · Short answer* How many terms of the A.P. $27, 24, 21, \dots$ must be taken so that their sum is $105$ ? Which term of the A.P. is zero ? **Solution** 1. Given A.P. is $27, 24, 21, \dots$, so first term $a = 27$ and common difference $d = 24 - 27 = -3$. 2. Let the sum of $n$ terms be $105$. Using $S_n = \frac{n}{2}[2a + (n-1)d]$, we get $105 = \frac{n}{2}[2(27) + (n-1)(-3)]$. 3. Simplifying, $210 = n[54 - 3n + 3] = n[57 - 3n]$, which gives $3n^2 - 57n + 210 = 0$ or $n^2 - 19n + 70 = 0$. 4. Solving $n^2 - 14n - 5n + 70 = 0$, we get $(n - 14)(n - 5) = 0$, so $n = 5$ or $n = 14$. Both values are admissible. 5. For the second part, let the $k$-th term be zero: $a_k = a + (k-1)d = 0$. 6. Substituting the values, $27 + (k-1)(-3) = 0$, which gives $27 = 3(k-1)$ or $k - 1 = 9$, so $k = 10$. 7. Result: $5$ or $14$ terms must be taken, and the $10^{\text{th}}$ term is zero. **Answer:** $n = 5 \text{ or } 14$, and $10^{\text{th}}$ term is zero > Common mistake: Rejecting one of the two values of $n$ without checking both sums. ## CBSE Class 10 Maths Standard Question Paper 2024 (Set 30/1/1) with Solutions ### Question 4 *1 mark · MCQ* In an A.P., if the first term $a = 7$, $n$th term $a_n = 84$ and the sum of first $n$ terms $s_n = \frac{2093}{2}$, then $n$ is equal to : - $22$ - $24$ - $23$ - $26$ **Solution** 1. The sum of $n$ terms of an A.P. is given by $S_n = \frac{n}{2}(a + a_n)$. 2. Substituting the given values, $\frac{2093}{2} = \frac{n}{2}(7 + 84)$, which gives $2093 = n(91)$ and $n = \frac{2093}{91} = 23$. **Answer:** (c) $23$ > Common mistake: Students may try to use the formula $S_n = \frac{n}{2}[2a + (n-1)d]$ and get stuck due to the unknown common difference $d$. ### Question 33 *5 marks · Long answer* The sum of first and eighth terms of an A.P. is $32$ and their product is $60$. Find the first term and common difference of the A.P. Hence, also find the sum of its first $20$ terms. **Solution** 1. Given $a_1 + a_8 = 32$, which implies $a + (a + 7d) = 32$, so $2a + 7d = 32$. 2. Given $a_1 \times a_8 = 60$, which implies $a(a + 7d) = 60$. 3. Substitute $2a + 7d = 32$ or express terms: since $a_1 + a_8 = 32$ and product is $60$, let the two terms be $x$ and $y$ where $x+y=32$ and $xy=60$. 4. Solving $z^2 - 32z + 60 = 0$, we get $(z - 30)(z - 2) = 0$, so the first and eighth terms are $2$ and $30$ (or vice versa). 5. Case 1: $a = 2$ and $a_8 = 30$. Since $a + 7d = 30$, we get $2 + 7d = 30$, so $7d = 28$, giving $d = 4$. 6. Case 2: $a = 30$ and $a_8 = 2$. Since $a + 7d = 2$, we get $30 + 7d = 2$, so $7d = -28$, giving $d = -4$. 7. For $a = 2, d = 4$, the sum of first $20$ terms is $S_{20} = \frac{20}{2} [2(2) + (20-1)(4)] = 10 [4 + 76] = 800$. 8. For $a = 30, d = -4$, the sum of first $20$ terms is $S_{20} = \frac{20}{2} [2(30) + (20-1)(-4)] = 10 [60 - 76] = -160$. **Answer:** First term $2$, common difference $4$, sum of $20$ terms $800$ (or first term $30$, common difference $-4$, sum $-160$). > Common mistake: Failing to consider both possible cases for first term and common difference. ### Question 33 (OR) *5 marks · Long answer* In an A.P. of $40$ terms, the sum of first $9$ terms is $153$ and the sum of last $6$ terms is $687$. Determine the first term and common difference of A.P. Also, find the sum of all the terms of the A.P. **Solution** 1. Given an A.P. of $n = 40$ terms. Sum of first $9$ terms is $S_9 = 153$, so $\frac{9}{2}[2a + 8d] = 153$, which gives $2a + 8d = 153 \times \frac{2}{9} = 34$, or $a + 4d = 17$ (Equation 1). 2. The sum of the last $6$ terms is the sum of all $40$ terms minus the sum of the first $34$ terms: $S_{40} - S_{34} = 687$. 3. Alternatively, the last $6$ terms start from the $35^{\text{th}}$ term: $a_{35} + a_{36} + a_{37} + a_{38} + a_{39} + a_{40} = 687$. 4. Using $a_n = a + (n-1)d$, this gives $(a + 34d) + (a + 35d) + (a + 36d) + (a + 37d) + (a + 38d) + (a + 39d) = 687$, which simplifies to $6a + 219d = 687$, or $2a + 73d = 229$ (Equation 2). 5. Multiply Equation 1 by 2: $2a + 8d = 34$. Subtract from Equation 2: $(2a + 73d) - (2a + 8d) = 229 - 34$, so $65d = 195$, giving $d = 3$. 6. Substitute $d = 3$ into Equation 1: $a + 4(3) = 17$, so $a = 5$. 7. Find the sum of all $40$ terms: $S_{40} = \frac{40}{2} [2(5) + (40-1)(3)] = 20 [10 + 39 \times 3] = 20 [10 + 117] = 20 \times 127 = 2540$. **Answer:** First term $5$, common difference $3$, sum of all terms $2540$. > Common mistake: Errors in writing the sum of the last few terms. ## CBSE Class 10 Maths Basic Question Paper 2023 (Set 430/1/1) with Solutions ### Question 6 *1 mark · MCQ* The seventh term of an A.P. whose first term is $28$ and common difference $-4$, is - $0$ - $4$ - $52$ - $56$ **Solution** 1. Given first term $a = 28$ and common difference $d = -4$. 2. The $n$-th term of an A.P. is given by $a_n = a + (n - 1)d$. 3. The seventh term is $a_7 = 28 + (7 - 1)(-4) = 28 + 6(-4) = 28 - 24 = 4$. **Answer:** (b) $4$ > Common mistake: Using $7d$ instead of $(7-1)d$ in the formula. ### Question 36 *4 marks · Case-based* Aahana being a plant lover decides to convert her balcony into beautiful garden full of plants. She bought few plants with pots for her balcony. She placed the pots in such a way that number of pots in the first row is $2$, second row is $5$, third row is $8$ and so on. Based on the above information, answer the following questions : (i) Find the number of pots placed in the $10^{\text{th}}$ row. (ii) Find the difference in the number of pots placed in $5^{\text{th}}$ row and $2^{\text{nd}}$ row. (iii) If Aahana wants to place $100$ pots in total, then find the total number of rows formed in the arrangement. **Part (i)** 1. The number of pots in each row forms an arithmetic progression: $2, 5, 8, \dots$. 2. Here, first term $a = 2$ and common difference $d = 5 - 2 = 3$. 3. Using the formula $a_n = a + (n - 1)d$, the number of pots in the $10^{\text{th}}$ row is $a_{10} = 2 + (10 - 1)3 = 2 + 27 = 29$. Answer (i): $29$ **Part (ii)** 1. The number of pots in the $5^{\text{th}}$ row is $a_5 = 2 + (5 - 1)3 = 2 + 12 = 14$. 2. The number of pots in the $2^{\text{nd}}$ row is $a_2 = 5$. 3. The difference is $14 - 5 = 9$. Answer (ii): $9$ **Part (iii)** 1. Let the total number of rows be $n$. Given the sum of $n$ terms $S_n = 100$. 2. Using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$, we get $100 = \frac{n}{2}[2(2) + (n - 1)3]$. 3. Simplifying gives $200 = n(4 + 3n - 3) = n(3n + 1)$, which leads to the quadratic equation $3n^2 + n - 200 = 0$. 4. Solving $(3n + 25)(n - 8) = 0$, we get $n = 8$ (since $n$ cannot be negative). Answer (iii): $8$ **Answer:** Total rows: 8 > Common mistake: Confusing the term value $a_n$ with the sum of terms $S_n$. ### Question 36 (OR) *4 marks · Case-based* (iii) If Aahana has sufficient space for $12$ rows, then how many total number of pots are placed by her with the same arrangement ? **Part (iii)** 1. Given the number of rows $n = 12$, first term $a = 2$, and common difference $d = 3$. 2. Using the sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$, we find the total number of pots. 3. Substitute the values: $S_{12} = \frac{12}{2}[2(2) + (12 - 1)3] = 6[4 + 33] = 6 \times 37 = 222$. Answer (iii): $222$ **Answer:** Total pots: 222 > Common mistake: Calculation error while multiplying 6 and 37. ## CBSE Class 10 Maths Standard Question Paper 2023 (Set 30/1/1) with Solutions ### Question 3 *1 mark · MCQ* If $p - 1, p + 1$ and $2p + 3$ are in A.P., then the value of $p$ is - -2 - 4 - 0 - 2 **Solution** 1. If $p - 1, p + 1, 2p + 3$ are in A.P., the difference between consecutive terms is equal, so $(p + 1) - (p - 1) = (2p + 3) - (p + 1)$. 2. Simplify both sides: $2 = p + 2$. 3. Solve for $p$: $p = 2 - 2 = 0$. 4. Checking the options, $p = 0$ corresponds to option (c). **Answer:** (c) 0 > Common mistake: Subtracting terms in the wrong order or setting sums equal instead of differences. ### Question 35 *5 marks · Long answer* The ratio of the $11^{\text{th}}$ term to $17^{\text{th}}$ term of an A.P. is $3 : 4$. Find the ratio of $5^{\text{th}}$ term to $21^{\text{st}}$ term of the same A.P. Also, find the ratio of the sum of first 5 terms to that of first 21 terms. **Solution** 1. Let the first term of the A.P. be $a$ and the common difference be $d$. 2. Given that the ratio of the $11^{\text{th}}$ term to the $17^{\text{th}}$ term is $3 : 4$, so $\frac{a + 10d}{a + 16d} = \frac{3}{4}$. 3. Cross-multiplying gives $4(a + 10d) = 3(a + 16d) \implies 4a + 40d = 3a + 48d \implies a = 8d$. 4. We need to find the ratio of the $5^{\text{th}}$ term to the $21^{\text{st}}$ term: $\frac{a_5}{a_{21}} = \frac{a + 4d}{a + 20d}$. 5. Substitute $a = 8d$: $\frac{8d + 4d}{8d + 20d} = \frac{12d}{28d} = \frac{12}{28} = \frac{3}{7}$. 6. Now, find the ratio of the sum of the first 5 terms to that of the first 21 terms: $\frac{S_5}{S_{21}} = \frac{\frac{5}{2}[2a + 4d]}{\frac{21}{2}[2a + 20d]} = \frac{5(2a + 4d)}{21(2a + 20d)}$. 7. Substitute $a = 8d$ into the sums ratio: $\frac{5(2(8d) + 4d)}{21(2(8d) + 20d)} = \frac{5(16d + 4d)}{21(16d + 20d)} = \frac{5(20d)}{21(36d)} = \frac{100}{756}$. 8. Simplifying the fraction $\frac{100}{756}$ by dividing by $4$ gives $\frac{25}{189}$. **Answer:** Ratio of 5th to 21st term = $3 : 7$, Ratio of sum of first 5 to 21 terms = $25 : 189$ > Common mistake: Applying the formula for the $n^{\text{th}}$ term incorrectly as $a + nd$ instead of $a + (n-1)d$. ### Question 35 (OR) *5 marks · Long answer* 250 logs are stacked in the following manner : 22 logs in the bottom row, 21 in the next row, 20 in the row next to it and so on (as shown by an example). In how many rows, are the 250 logs placed and how many logs are there in the top row? **Solution** 1. The number of logs in each row forms an Arithmetic Progression: $22, 21, 20, \dots$ 2. Here, the first term $a = 22$, the common difference $d = 21 - 22 = -1$, and the total number of logs $S_n = 250$. 3. Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$, substitute the known values: $250 = \frac{n}{2}[2(22) + (n-1)(-1)]$. 4. Simplify the equation: $500 = n[44 - n + 1] \implies 500 = n[45 - n] \implies n^2 - 45n + 500 = 0$. 5. Factorize the quadratic equation: $n^2 - 25n - 20n + 500 = 0 \implies n(n - 25) - 20(n - 25) = 0$. 6. This gives $(n - 25)(n - 20) = 0$, so $n = 25$ or $n = 20$. 7. If $n = 25$, the number of logs in the $25^{\text{th}}$ row is $a_{25} = a + 24d = 22 + 24(-1) = -2$, which is not possible since the number of logs cannot be negative. 8. Therefore, $n = 20$. 9. The number of logs in the top row ($20^{\text{th}}$ row) is $a_{20} = a + 19d = 22 + 19(-1) = 22 - 19 = 3$. **Answer:** Number of rows = $20$, Number of logs in the top row = $3$ > Common mistake: Accepting both values of $n$ without checking if the number of logs in the top row becomes negative. ## CBSE Class 10 Maths Basic Question Paper 2022 (Set 430/1/1) with Solutions ### Question 1 (a) *2 marks · Short answer* In an AP, if $a = 50$, $d = -4$ and $S_n = 0$, then find the value of $n$. **Solution** 1. Use the formula for the sum of $n$ terms of an AP: $S_n = \frac{n}{2}[2a + (n-1)d]$. 2. Substitute the given values $a = 50$, $d = -4$, and $S_n = 0$: $0 = \frac{n}{2}[2(50) + (n-1)(-4)]$. 3. Simplify the equation: $100 - 4n + 4 = 0$, which gives $104 - 4n = 0$. 4. Solve for $n$: $4n = 104$, so $n = 26$. **Answer:** $n = 26$ > Common mistake: Forgetting that $n$ must be a positive integer or making sign errors while expanding the bracket. ### Question 1 (OR) *2 marks · Short answer* Find the sum of the first twelve 2-digit multiples of $7$, using an AP. **Solution** 1. The first twelve 2-digit multiples of $7$ form an AP: $14, 21, 28, \dots$ up to $12$ terms. 2. Here, the first term $a = 14$, the common difference $d = 7$, and the number of terms $n = 12$. 3. Use the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$ and substitute the values: $S_{12} = \frac{12}{2}[2(14) + (12-1)7]$. 4. Calculate the value: $S_{12} = 6[28 + 77] = 6[105] = 630$. **Answer:** $630$ > Common mistake: Starting the AP from $7$ instead of the first 2-digit multiple which is $14$. ### Question 13 (a) *2 marks · Case-based* Based on the given information about old clothes and a footmat (rug) where stitches in circular rows make a pattern: $6, 12, 18, 24, \dots$ Check whether the given pattern forms an AP. If yes, find the common difference and the next term of the AP. **Part (i)** 1. Given sequence of stitches is $6, 12, 18, 24, \dots$ 2. Check the difference between consecutive terms: $a_2 - a_1 = 12 - 6 = 6$, $a_3 - a_2 = 18 - 12 = 6$, $a_4 - a_3 = 24 - 18 = 6$. 3. Since the difference between each term and its preceding term is constant ($d = 6$), the given pattern forms an AP. Answer (i): Yes, it forms an AP with common difference $d = 6$. **Part (ii)** 1. The next term is the fifth term of the AP ($a_5$). 2. $a_5 = a_4 + d = 24 + 6 = 30$. Answer (ii): The next term of the AP is 30. **Answer:** The pattern forms an AP with common difference 6 and next term 30. > Common mistake: Checking only one pair of consecutive terms instead of verifying multiple pairs. ### Question 13 (b) *2 marks · Case-based* Write the $n^{\text{th}}$ term of the AP. Hence, find the number of stitches in the $10^{\text{th}}$ circular row. **Part (i)** 1. The given AP has first term $a = 6$ and common difference $d = 6$. 2. The $n^{\text{th}}$ term of an AP is given by $a_n = a + (n - 1)d$. 3. Substitute $a = 6$ and $d = 6$: $a_n = 6 + (n - 1)6 = 6 + 6n - 6 = 6n$. Answer (i): The $n^{\text{th}}$ term is $6n$. **Part (ii)** 1. To find the number of stitches in the $10^{\text{th}}$ row, substitute $n = 10$ in the $n^{\text{th}}$ term formula. 2. $a_{10} = 6(10) = 60$. Answer (ii): The number of stitches in the $10^{\text{th}}$ row is 60. **Answer:** The nth term is $6n$ and the 10th term is 60. > Common mistake: Multiplying incorrectly or confusing $n$ with the common difference. ## CBSE Class 10 Maths Standard Question Paper 2022 (Set 30/1/1) with Solutions ### Question 1 (a) *2 marks · Short answer* Find the sum of first 30 terms of AP : $-30, -24, -18, \dots \dots$. **Solution** 1. Identify the first term $a = -30$ and common difference $d = -24 - (-30) = 6$. 2. Use the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$ with $n = 30$. 3. Substitute the values: $S_{30} = \frac{30}{2}[2(-30) + (30-1)(6)]$. 4. Calculate the result: $S_{30} = 15[-60 + 174] = 15(114) = 1710$. **Answer:** $1710$ > Common mistake: Arithmetic error in finding the common difference $d$ or calculation inside the bracket. ### Question 1 (b) (OR) *2 marks · Short answer* In an AP if $S_n = n(4n + 1)$, then find the AP. **Solution** 1. Use the relation between $n$-th term and sum of $n$ terms: $a_n = S_n - S_{n-1}$. 2. Find $S_1 = 1(4(1) + 1) = 5$, which is the first term $a_1$. 3. Find $S_2 = 2(4(2) + 1) = 18$, so the second term $a_2 = S_2 - S_1 = 18 - 5 = 13$. 4. Find the common difference $d = a_2 - a_1 = 13 - 5 = 8$, thus the AP is $5, 13, 21, \dots$. **Answer:** $5, 13, 21, \dots$ > Common mistake: Confusing $S_n$ with $a_n$. ### Question 13 (a) *2 marks · Case-based* (Case Study 1) Write the AP for the number of triangles used in the figures. Also, write the $n^{\text{th}}$ term of this AP. **Part (i)** 1. Observe the number of triangles in successive figures: 3, 5, 7, and so on. 2. This forms an AP with first term $a = 3$ and common difference $d = 5 - 3 = 2$. Answer (i): AP: $3, 5, 7, \dots$ **Part (ii)** 1. Use the formula for the $n^{\text{th}}$ term of an AP: $a_n = a + (n - 1)d$. 2. Substitute $a = 3$ and $d = 2$: $a_n = 3 + (n - 1)2 = 3 + 2n - 2 = 2n + 1$. Answer (ii): $a_n = 2n + 1$ **Answer:** AP is $3, 5, 7, \dots$ and the $n^{\text{th}}$ term is $2n + 1$. > Common mistake: Writing the common difference incorrectly as 1 instead of 2. ### Question 13 (b) *2 marks · Case-based* (Case Study 1) Which figure has $61$ matchsticks? **Part (i)** 1. Observe the matchstick pattern where the first figure has 4 matchsticks, the second has 7, and the third has 10, forming an arithmetic progression. 2. Identify the first term $a = 4$ and the common difference $d = 7 - 4 = 3$. 3. Use the nth term formula $a_n = a + (n - 1)d$ and substitute $a_n = 61$ to find $n$. 4. Solve $61 = 4 + (n - 1)3$, which gives $57 = 3(n - 1)$, so $n - 1 = 19$ and $n = 20$. Answer (i): Figure 20 **Answer:** The 20th figure has 61 matchsticks. > Common mistake: Confusing the term number with the common difference or miscounting the initial number of matchsticks. ## Related pages - [Arithmetic Progressions: NCERT solutions](https://www.swavid.com/maths/class/10/chapter/arithmetic-progressions/ncert-solutions) - [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student