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Sequences and Series: JEE Main Previous Year Questions (2021-2026)
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Page 1 JEE Main Previous Year Questions (2021-2026): Sequences and Series (January 2026) Q1: Let the arithmetic mean of 1/a and 1/b be 5/16, a > 2. If a is such that a, 4, a, b are in A.P., then the equation ax² - ax + 2(a - 2b) = 0 has : (a) one root in (1, 4) and another in (-2, 0) (b) one root in (0, 2) and another in (-4, -2) (c) both roots in the interval (-2, 0) (d) complex roots of magnitude less than 2 Ans: (a) Sol: The arithmetic mean of 1/a and 1/b be 5/16 a, 4, a, b are in A.P. Let the common difference be d. (A). 4 = a + d ? d = 4 - a (B). a = a + 2d = a + 2(4 - a) = 8 - a (C). b = a + 3d = a + 3(4 - a) = 12 - 2a Solve for a, b, and a Substitute b = 12 - 2a into equation (1) : 8(a + 12 - 2a) = 5a(12 - 2a) 8(12 - a) = 60a - 10a² 96 - 8a = 60a - 10a² 10a² - 68a + 96 = 0 5a² - 34a + 48 = 0 Page 2 JEE Main Previous Year Questions (2021-2026): Sequences and Series (January 2026) Q1: Let the arithmetic mean of 1/a and 1/b be 5/16, a > 2. If a is such that a, 4, a, b are in A.P., then the equation ax² - ax + 2(a - 2b) = 0 has : (a) one root in (1, 4) and another in (-2, 0) (b) one root in (0, 2) and another in (-4, -2) (c) both roots in the interval (-2, 0) (d) complex roots of magnitude less than 2 Ans: (a) Sol: The arithmetic mean of 1/a and 1/b be 5/16 a, 4, a, b are in A.P. Let the common difference be d. (A). 4 = a + d ? d = 4 - a (B). a = a + 2d = a + 2(4 - a) = 8 - a (C). b = a + 3d = a + 3(4 - a) = 12 - 2a Solve for a, b, and a Substitute b = 12 - 2a into equation (1) : 8(a + 12 - 2a) = 5a(12 - 2a) 8(12 - a) = 60a - 10a² 96 - 8a = 60a - 10a² 10a² - 68a + 96 = 0 5a² - 34a + 48 = 0 Using the quadratic formula: a = 48/10 = 4.8, a = 20/10 = 2 (Discarded because the problem states a > 2) With a = 4.8 : a = 8 - 4.8 = 3.2, b = 12 - 2(4.8) = 12 - 9.6 = 2.4 Analyze the Quadratic Equation The given equation is ax² - ax + 2(a - 2b) = 0. Substitute the values: 3.2x² - 4.8x + 2(3.2 - 2(2.4)) = 0 3.2x² - 4.8x + 2(3.2 - 4.8) = 0 3.2x² - 4.8x - 3.2 = 0 2x² - 3x - 2 = 0 Factorise (2x + 1)(x - 2) = 0 The roots are x = 2 and x = -0.5. Looking at the options: Root x = 2 lies in the interval (1, 4). Root x = -0.5 lies in the interval (-2, 0). Q2: is equal to : (a) 2 26 (b) 3 25 (c) 3 26 (d) 2 25 Ans: (a) Sol: Let Notice that from the second term onwards, the terms follow a pattern. Let S' be the sum of these terms: Page 3 JEE Main Previous Year Questions (2021-2026): Sequences and Series (January 2026) Q1: Let the arithmetic mean of 1/a and 1/b be 5/16, a > 2. If a is such that a, 4, a, b are in A.P., then the equation ax² - ax + 2(a - 2b) = 0 has : (a) one root in (1, 4) and another in (-2, 0) (b) one root in (0, 2) and another in (-4, -2) (c) both roots in the interval (-2, 0) (d) complex roots of magnitude less than 2 Ans: (a) Sol: The arithmetic mean of 1/a and 1/b be 5/16 a, 4, a, b are in A.P. Let the common difference be d. (A). 4 = a + d ? d = 4 - a (B). a = a + 2d = a + 2(4 - a) = 8 - a (C). b = a + 3d = a + 3(4 - a) = 12 - 2a Solve for a, b, and a Substitute b = 12 - 2a into equation (1) : 8(a + 12 - 2a) = 5a(12 - 2a) 8(12 - a) = 60a - 10a² 96 - 8a = 60a - 10a² 10a² - 68a + 96 = 0 5a² - 34a + 48 = 0 Using the quadratic formula: a = 48/10 = 4.8, a = 20/10 = 2 (Discarded because the problem states a > 2) With a = 4.8 : a = 8 - 4.8 = 3.2, b = 12 - 2(4.8) = 12 - 9.6 = 2.4 Analyze the Quadratic Equation The given equation is ax² - ax + 2(a - 2b) = 0. Substitute the values: 3.2x² - 4.8x + 2(3.2 - 2(2.4)) = 0 3.2x² - 4.8x + 2(3.2 - 4.8) = 0 3.2x² - 4.8x - 3.2 = 0 2x² - 3x - 2 = 0 Factorise (2x + 1)(x - 2) = 0 The roots are x = 2 and x = -0.5. Looking at the options: Root x = 2 lies in the interval (1, 4). Root x = -0.5 lies in the interval (-2, 0). Q2: is equal to : (a) 2 26 (b) 3 25 (c) 3 26 (d) 2 25 Ans: (a) Sol: Let Notice that from the second term onwards, the terms follow a pattern. Let S' be the sum of these terms: Calculate the sum of the GP The series is a geometric progression with first term a = 1, common ratio r = 6, and n = 25 terms. The sum formula is Substituting this back into S': Find the total sum S: Now, add the first term 6/3 26 back to S': Page 4 JEE Main Previous Year Questions (2021-2026): Sequences and Series (January 2026) Q1: Let the arithmetic mean of 1/a and 1/b be 5/16, a > 2. If a is such that a, 4, a, b are in A.P., then the equation ax² - ax + 2(a - 2b) = 0 has : (a) one root in (1, 4) and another in (-2, 0) (b) one root in (0, 2) and another in (-4, -2) (c) both roots in the interval (-2, 0) (d) complex roots of magnitude less than 2 Ans: (a) Sol: The arithmetic mean of 1/a and 1/b be 5/16 a, 4, a, b are in A.P. Let the common difference be d. (A). 4 = a + d ? d = 4 - a (B). a = a + 2d = a + 2(4 - a) = 8 - a (C). b = a + 3d = a + 3(4 - a) = 12 - 2a Solve for a, b, and a Substitute b = 12 - 2a into equation (1) : 8(a + 12 - 2a) = 5a(12 - 2a) 8(12 - a) = 60a - 10a² 96 - 8a = 60a - 10a² 10a² - 68a + 96 = 0 5a² - 34a + 48 = 0 Using the quadratic formula: a = 48/10 = 4.8, a = 20/10 = 2 (Discarded because the problem states a > 2) With a = 4.8 : a = 8 - 4.8 = 3.2, b = 12 - 2(4.8) = 12 - 9.6 = 2.4 Analyze the Quadratic Equation The given equation is ax² - ax + 2(a - 2b) = 0. Substitute the values: 3.2x² - 4.8x + 2(3.2 - 2(2.4)) = 0 3.2x² - 4.8x + 2(3.2 - 4.8) = 0 3.2x² - 4.8x - 3.2 = 0 2x² - 3x - 2 = 0 Factorise (2x + 1)(x - 2) = 0 The roots are x = 2 and x = -0.5. Looking at the options: Root x = 2 lies in the interval (1, 4). Root x = -0.5 lies in the interval (-2, 0). Q2: is equal to : (a) 2 26 (b) 3 25 (c) 3 26 (d) 2 25 Ans: (a) Sol: Let Notice that from the second term onwards, the terms follow a pattern. Let S' be the sum of these terms: Calculate the sum of the GP The series is a geometric progression with first term a = 1, common ratio r = 6, and n = 25 terms. The sum formula is Substituting this back into S': Find the total sum S: Now, add the first term 6/3 26 back to S': Q3: The value of is (a) e/2 (b) ve (c) 2/e (d) 1/e Ans: (d) Sol: First, let's simplify the expression inside the summation. We can cancel out the k in the numerator with the k in k!: Now, we rewrite the numerator (k + 1) in terms of (k - 1) to further simplify the fraction: k + 1 = (k - 1) + 2 Substituting this back into the expression: For the first part, (valid for k = 2). So, the general term a k becomes: Expand the Series Page 5 JEE Main Previous Year Questions (2021-2026): Sequences and Series (January 2026) Q1: Let the arithmetic mean of 1/a and 1/b be 5/16, a > 2. If a is such that a, 4, a, b are in A.P., then the equation ax² - ax + 2(a - 2b) = 0 has : (a) one root in (1, 4) and another in (-2, 0) (b) one root in (0, 2) and another in (-4, -2) (c) both roots in the interval (-2, 0) (d) complex roots of magnitude less than 2 Ans: (a) Sol: The arithmetic mean of 1/a and 1/b be 5/16 a, 4, a, b are in A.P. Let the common difference be d. (A). 4 = a + d ? d = 4 - a (B). a = a + 2d = a + 2(4 - a) = 8 - a (C). b = a + 3d = a + 3(4 - a) = 12 - 2a Solve for a, b, and a Substitute b = 12 - 2a into equation (1) : 8(a + 12 - 2a) = 5a(12 - 2a) 8(12 - a) = 60a - 10a² 96 - 8a = 60a - 10a² 10a² - 68a + 96 = 0 5a² - 34a + 48 = 0 Using the quadratic formula: a = 48/10 = 4.8, a = 20/10 = 2 (Discarded because the problem states a > 2) With a = 4.8 : a = 8 - 4.8 = 3.2, b = 12 - 2(4.8) = 12 - 9.6 = 2.4 Analyze the Quadratic Equation The given equation is ax² - ax + 2(a - 2b) = 0. Substitute the values: 3.2x² - 4.8x + 2(3.2 - 2(2.4)) = 0 3.2x² - 4.8x + 2(3.2 - 4.8) = 0 3.2x² - 4.8x - 3.2 = 0 2x² - 3x - 2 = 0 Factorise (2x + 1)(x - 2) = 0 The roots are x = 2 and x = -0.5. Looking at the options: Root x = 2 lies in the interval (1, 4). Root x = -0.5 lies in the interval (-2, 0). Q2: is equal to : (a) 2 26 (b) 3 25 (c) 3 26 (d) 2 25 Ans: (a) Sol: Let Notice that from the second term onwards, the terms follow a pattern. Let S' be the sum of these terms: Calculate the sum of the GP The series is a geometric progression with first term a = 1, common ratio r = 6, and n = 25 terms. The sum formula is Substituting this back into S': Find the total sum S: Now, add the first term 6/3 26 back to S': Q3: The value of is (a) e/2 (b) ve (c) 2/e (d) 1/e Ans: (d) Sol: First, let's simplify the expression inside the summation. We can cancel out the k in the numerator with the k in k!: Now, we rewrite the numerator (k + 1) in terms of (k - 1) to further simplify the fraction: k + 1 = (k - 1) + 2 Substituting this back into the expression: For the first part, (valid for k = 2). So, the general term a k becomes: Expand the Series Let's apply the summation to both parts. Note that for k = 1, the term 1 / (k - 2)! is effectively 0 because the factorial of a negative integer is infinite in the denominator. Using exponential expansion(Taylor series expansion for e x :) Evaluate the Summations Let n = k - 2. When k = 2, n = 0. When k ? 8, n ? 8. The exponent becomes k + 1 = (n + 2) + 1 = n + 3. Let m = k - 1. When k = 1, m = 0. When k ? 8, m ? 8. The exponent becomes k + 1 = (m + 1) + 1 = m + 2. Required SRead More
FAQs on Sequences and Series: JEE Main Previous Year Questions (2021-2026)
| 1. How do I identify whether a sequence is arithmetic or geometric in JEE Main questions? |  |
Ans. An arithmetic sequence has a constant common difference (d) between consecutive terms, while a geometric sequence has a constant common ratio (r). Check if the difference between consecutive terms is fixed (arithmetic) or if each term divided by the previous term is constant (geometric). Previous year JEE Main questions often test this distinction in mixed-series problems.
| 2. What's the fastest way to find the sum of an arithmetic progression when the number of terms is large? | |
Ans. Use the formula S_n = n/2 × (first term + last term) or S_n = n/2 × [2a + (n-1)d], where n is the number of terms and d is the common difference. This avoids calculating individual terms and appears frequently in JEE Main arithmetic series problems requiring quick computation under time pressure.
| 3. Why do some geometric series converge while others diverge in JEE problems? | |
Ans. A geometric series converges only when the common ratio's absolute value is less than 1 (|r| < 1). If |r| ≥ 1, the series diverges to infinity. JEE Main questions test this concept through infinite series summation, where identifying convergence is essential before applying the sum formula S = a/(1-r).
| 4. How do I spot when to use the arithmetic-geometric series formula in JEE Main previous year papers? | |
Ans. Arithmetic-geometric series combine arithmetic and geometric patterns-terms form an AP while coefficients follow a GP pattern. Look for series like a + (a+d)r + (a+2d)r² + ... in JEE Main questions. Recognizing this mixed structure immediately signals using the special AG series formula rather than standard AP or GP methods.
| 5. What are the most common mistakes students make with summation notation in JEE Main sequences and series problems? | |
Ans. Students often misinterpret sigma notation limits, confuse index variables, or incorrectly apply formulas for non-standard ranges. A frequent error involves applying closed-form AP or GP formulas to sequences starting at indices other than 1. Careful index tracking and verifying formula applicability prevents these costly mistakes in JEE Main exams.
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