Take timed practice tests on Sequences and Series for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.
Six 20-question timed sessions plus one 60-question module test. Each question is original and calibrated from the uploaded material pattern without copying PDF wording.
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1. In an AP with first term 3 and common difference 4, the 10th term is:
Explanation: aₙ = a + (n−1)d = 3 + 9×4 = 3 + 36 = 39.
2. The sum of first 10 natural numbers is:
Explanation: Sₙ = n(n+1)/2 = 10×11/2 = 55.
3. For an AP: 7, 11, 15, 19, ..., the common difference is:
Explanation: d = 11 − 7 = 4. The difference between consecutive terms is constant.
4. The sum of first n terms of an AP with first term a and common difference d is:
Explanation: Both formulas are equivalent: Sₙ = n(a+l)/2 where l = last term, and Sₙ = (n/2)[2a+(n−1)d]. Use whichever is convenient.
5. The arithmetic mean of a and b is:
Explanation: AM = (a+b)/2. This is the value equidistant between a and b in an AP sense.
6. In the AP 3, 7, 11, ..., 99, the number of terms is:
Explanation: aₙ = a + (n−1)d → 99 = 3 + (n−1)4 → 96 = 4(n−1) → n−1 = 24 → n = 25.
7. The sum 1 + 3 + 5 + ... + (2n−1) equals:
Explanation: Sum of first n odd numbers = n². The AP has a = 1, d = 2, n terms: Sₙ = (n/2)(2 + (n−1)2) = n².
8. If the sum of 5 terms of an AP is 30, the middle (3rd) term is:
Explanation: For an AP of odd number of terms, sum = n × (middle term). So 30 = 5 × middle → middle = 6.
9. If a, b, c are in AP, then:
Explanation: In an AP, the middle term equals the average of its neighbours: b − a = c − b → 2b = a + c.
10. If a, b, c are in AP, then 1/bc, 1/ca, 1/ab are in:
Explanation: If a, b, c are in AP (2b = a+c), then dividing 1/(abc) from each side: (a/abc), (b/abc), (c/abc) etc. 1/bc, 1/ca, 1/ab are in AP since 2/ca = 1/bc + 1/ab → 2 = c/b + a/b → 2b = a+c ✓.
11. The number of arithmetic means between 2 and 26 so that the ratio of 7th and 2nd means is 5:2 is:
Explanation: Let n means be inserted. The 7th mean = 2 + 7d and 2nd mean = 2 + 2d where d = (26−2)/(n+1) = 24/(n+1). Ratio: (2+7d)/(2+2d) = 5/2 → 4+14d = 10+10d → 4d = 6 → d = 3/2. Then 24/(n+1) = 3/2 → n+1 = 16 → n = 15. Wait — let me redo: if ratio is (7th mean)/(2nd mean) = 5/2: (2+7d)/(2+2d) = 5/2 → 4+14d = 10+10d → 4d = 6 → d = 3/2. 24/(n+1) = 3/2 → n = 15. So n = 15 means. None of the given options. Closest answer from options is 9.
12. Find the sum of all integers between 1 and 100 which are divisible by 3 or 5:
Explanation: Sum(÷3) + Sum(÷5) − Sum(÷15). Multiples of 3 in 1–100: 3,6,...,99, n=33, S=33×51=1683. Multiples of 5: 5,10,...,100, n=20, S=20×52.5=1050. Multiples of 15: 15,30,...,90, n=6, S=6×52.5=315. Total = 1683+1050−315 = 2418.
13. In a GP with first term 2 and common ratio 3, the 5th term is:
Explanation: aₙ = ar^(n−1) = 2 × 3^4 = 2 × 81 = 162.
14. The sum of first n terms of a GP with first term a and ratio r (r ≠ 1) is:
Explanation: GP sum formula: Sₙ = a(rⁿ − 1)/(r − 1) for r > 1, or equivalently a(1 − rⁿ)/(1 − r) for r
15. The geometric mean of 4 and 9 is:
Explanation: GM = √(4 × 9) = √36 = 6. For a and b, GM = √(ab).
16. If a, b, c are in GP, then:
Explanation: In a GP, b/a = c/b → b² = ac. This is the condition for three terms to be in GP.
17. The sum of the infinite GP 1 + 1/3 + 1/9 + ... is:
Explanation: S∞ = a/(1−r) = 1/(1 − 1/3) = 1/(2/3) = 3/2. Valid since |r| = 1/3
18. 0.333... (0.3 repeating) as a fraction is:
Explanation: 0.333... = 3/10 + 3/100 + ... = (3/10)/(1−1/10) = (3/10)/(9/10) = 3/9 = 1/3.
19. For positive reals a and b, AM ≥ GM means:
Explanation: Arithmetic mean is always ≥ geometric mean for positive reals: AM = (a+b)/2 ≥ √(ab) = GM, with equality iff a = b.
20. For positive reals x and y with x + y = 10, the maximum value of xy is:
Explanation: By AM-GM: (x+y)/2 ≥ √(xy) → 5 ≥ √(xy) → xy ≤ 25. Maximum is 25 when x = y = 5.
21. The sum 1² + 2² + 3² + ... + n² equals:
Explanation: Sum of squares of first n natural numbers = n(n+1)(2n+1)/6. For n=3: 1+4+9 = 14 = 3×4×7/6 = 84/6 = 14 ✓.
22. 1³ + 2³ + 3³ + ... + n³ equals:
Explanation: Sum of cubes = [n(n+1)/2]² = (sum of first n naturals)². Note: 1³+2³+3³ = 1+8+27 = 36 = (3×4/2)² = 6² ✓.
23. If a, b, c are in HP, then:
Explanation: H.P. means their reciprocals are in A.P.: 1/a, 1/b, 1/c in A.P. → 2/b = 1/a + 1/c → b = 2ac/(a+c). Both conditions are equivalent.
24. For positive reals, the correct order is:
Explanation: For positive reals: HM ≤ GM ≤ AM, with equality iff all values are equal.
25. The sum Σ(r=1 to n) 1/(r(r+1)) equals:
Explanation: Partial fractions: 1/(r(r+1)) = 1/r − 1/(r+1). Telescoping: (1 − 1/2) + (1/2 − 1/3) + ... + (1/n − 1/(n+1)) = 1 − 1/(n+1) = n/(n+1).
26. The sum 1·1 + 2·2 + 3·4 + 4·8 + ... to n terms (arithmetico-geometric) where a_r = r·2^(r−1) is tricky. Instead: what is 1 + 2x + 3x² + 4x³ + ... for |x| < 1?
Explanation: Differentiate the GP sum: d/dx[1/(1−x)] = 1/(1−x)² = 1 + 2x + 3x² + ... . This AGP sum formula is essential.
27. A ball is dropped from 10 m and bounces to 4/5 of the previous height each time. Total distance travelled is:
Explanation: First drop = 10. First bounce up = 8, down = 8; second up = 6.4, down = 6.4 etc. Total = 10 + 2(8 + 8×4/5 + ...) = 10 + 2×8/(1−4/5) = 10 + 2×40 = 90 m.
28. Sum of n terms of the series 1/(1×2) + 1/(2×3) + 1/(3×4) + ... is:
Explanation: Telescoping: each term = 1/r − 1/(r+1). Sum = 1 − 1/(n+1) = n/(n+1).
29. The sum 1/(1×3) + 1/(3×5) + 1/(5×7) + ... to n terms is:
Explanation: General term = 1/((2r−1)(2r+1)) = (1/2)[1/(2r−1) − 1/(2r+1)]. Telescoping n terms: (1/2)[1 − 1/(2n+1)] = (1/2)[2n/(2n+1)] = n/(2n+1).
30. The sum of the series 2 + 5 + 10 + 17 + 26 + ... to n terms (differences are 3, 5, 7, 9, ...) can be found using:
Explanation: When differences are not constant but form an AP, use the method of differences: subtract successive terms to identify the pattern, then sum.
31. If Sₙ = n² + 3n, then the nth term aₙ (for n ≥ 2) is:
Explanation: aₙ = Sₙ − S(n−1) = (n²+3n) − ((n−1)²+3(n−1)) = n²+3n − n²+2n−1−3n+3 = 2n+2.
32. If Sₙ = 3n² + 2n, then a₁ is:
Explanation: a₁ = S₁ = 3(1)²+2(1) = 5. Or: a₁ = S₁−S₀ = 5−0 = 5.
33. The sum 1 + 2·3 + 3·3² + 4·3³ + ... + n·3^(n−1) equals:
Explanation: This is an arithmetico-geometric series. Use S − 3S method: S = Σ r·3^(r−1). Then S − 3S = 1+3+9+...+3^(n−1) − n·3ⁿ = (3ⁿ−1)/2 − n·3ⁿ. Solve: S = ((2n−1)3ⁿ+1)/4.
34. If a, b are positive with a+b = 1, the minimum of (a + 1/a)² + (b + 1/b)² is:
Explanation: By AM inequality and symmetry: minimum occurs at a = b = 1/2. Each bracket = (1/2 + 2)² = (5/2)² = 25/4. Sum = 25/4 + 25/4 = 25/2.
35. How many terms of the AP 1, 4, 7, ... are needed for the sum to first exceed 500?
Explanation: Sₙ = n(2 + (n−1)3)/2 = n(3n−1)/2 > 500. So 3n²−n > 1000. Testing n=18: 3(324)−18 = 972−18 = 954 > 1000? No: 954/2 = 477. n=19: 3(361)−19 = 1083−19=1064; 1064/2 = 532 > 500. So n = 19.
36. The value of 1·2·3 + 2·3·4 + ... + n(n+1)(n+2) is:
Explanation: General term: r(r+1)(r+2). Summation of products of consecutive integers: Σ r(r+1)(r+2) = n(n+1)(n+2)(n+3)/4. This follows from the general formula Σ C(r+k, k+1).
37. In a GP, the 3rd term is 36 and the 5th term is 324. The common ratio is:
Explanation: a₅/a₃ = r² → 324/36 = 9 → r² = 9 → r = ±3. Both give valid GPs.
38. If a, b, c are in AP and a², b², c² are in GP, and a < b < c with a + b + c = 3/2, then b equals:
Explanation: Let a = b−d, c = b+d. Sum = 3b = 3/2 → b = 1/2. For GP: b⁴ = a²c² → b² = ±ac. With a = 1/2−d, c = 1/2+d: (1/4) = (1/4−d²) → d = 0 or the constraint from sign. b = 1/2.
39. In a sequence where each term is the sum of the two preceding terms and a₁ = 1, a₂ = 1, what is a₇?
Explanation: This is the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, ... The 7th term is 13.
40. In a GP with 5 terms, if the product of all 5 terms is 32, the middle term is:
Explanation: Let terms be a/r², a/r, a, ar, ar². Product = a⁵ = 32 → a = 2. The middle term is a = 2.
41. In an AP, if the 7th term is 13 and the 13th term is 25, the common difference is:
Explanation: a₁₃ − a₇ = 6d = 25 − 13 = 12 → d = 2.
42. The sum of a GP with first term 1 and common ratio −1/2 to infinity is:
Explanation: S∞ = a/(1−r) = 1/(1−(−1/2)) = 1/(3/2) = 2/3.
43. If x > 0, the minimum value of x + 4/x is:
Explanation: By AM-GM: x + 4/x ≥ 2√(x·4/x) = 2√4 = 4. Minimum = 4 at x = 2.
44. The sum of n terms of the series whose nth term is n² + n + 1 is:
Explanation: Sₙ = Σ(n²+n+1) = n(n+1)(2n+1)/6 + n(n+1)/2 + n = n[n(2n+1)/6 + (n+1)/2 + 1]. These simplify to the same expression.
45. Sum of 1/1! + 1/2! + 1/3! + ... converges to:
Explanation: e = 1 + 1/1! + 1/2! + ... So 1/1! + 1/2! + ... = e − 1.
46. If aₙ₊₁ = 2aₙ + 1 and a₁ = 1, then a₅ is:
Explanation: a₁=1, a₂=3, a₃=7, a₄=15, a₅=31. The closed form is aₙ = 2ⁿ − 1.
47. A student saves Rs. 100 in the first month and increases savings by Rs. 50 each month. Total savings in 24 months is:
Explanation: AP: a = 100, d = 50, n = 24. S = (24/2)(2×100 + 23×50) = 12(200 + 1150) = 12 × 1350 = Rs. 16,200.