Let a_1 = 1 and for n geq 1 , a_n+1 = frac12 a_n + fracn^2 - 2n - 1n^2 (n + 1)^2 . Then left|sum_n=1^inftyleft(a_n - frac2n^2right)right| is equal

Numerical Answer Type:
Enter a numerical value Answer: 2 to 2 +4 marks

Solution & Explanation

### Related Formula Partial fraction decomposition for telescopic summing: frac2n^2 - (n+1)^2 + 1 dotsdots text structures directly cancel in series expansions. ### Core Logic Given recurrence: a_n+1 - frac12a_n = fracn^2 - 2n - 1n^2(n+1)^2 Rewrite the numerator to split the fraction: n^2 - 2n - 1 = 2n^2 - (n^2 + 2n + 1) = 2n^2 - (n+1)^2 a_n+1 - frac12a_n = frac2n^2 - (n+1)^2n^2(n+1)^2 = frac2(n+1)^2 - frac1n^2 ### Step 1: Telescope generation Multiply both sides by appropriate powers of 2 to create a cancelling chain: For n=1: a_2 - frac12a_1 = frac22^2 - frac11^2 For n=2: multiply by 2 Rightarrow 2left[a_3 - frac12a_2 = frac23^2 - frac12^2right] Rightarrow 2a_3 - a_2 = frac2 times 23^2 - frac22^2 Wait, let's look at a cleaner telescopic scaling: a_n+1 - frac2(n+1)^2 = frac12 left(a_n - frac2n^2right). ### Step 2: Identify Geometric Progression Let V_n = a_n - frac2n^2. The recurrence gives V_n+1 = frac12 V_n. This proves V_n is a geometric progression with common ratio r = 1/2. First term V_1 = a_1 - frac21^2 = 1 - 2 = -1. ### Step 3: Infinite Summation We need left| sum_n=1^infty left( a_n - frac2n^2 right) right| = left| sum_n=1^infty V_n right|. Since V_n is an infinite GP: S_infty = fracV_11 - r = frac-11 - 1/2 = frac-11/2 = -2 Taking absolute value: |-2| = 2 ### Pattern Recognition When dealing with rational fraction recurrences A_n+1 - k A_n = f(n) - k f(n-1), immediately substitute V_n = A_n - f(n). This substitution instantly isolates a classical Geometric Progression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

Reference Study Guides

Sequences And Series

More Sequences and Series Previous-Year Questions — Page 4

Q65 jee_main_2025_04_april_evening Arithmetic Progression
Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and q respectively. Let d and D be the common differences of AP's in A and B respectively such that mathrmD = mathrmd + 3, mathrm~d > 0. If fracmathrmp + mathrmqmathrmp - mathrmq = frac195, then mathrmp - mathrmq is equal to
  • A. 600
  • B. 450
  • C. 630
  • D. 540

Solution

### Core Logic Let the 3 elements of set A in A.P. be a-d, a, a+d. Their sum is 3a = 36 implies a = 12. Their product is p = a(a^2 - d^2) = 12(144 - d^2). Similarly, let the 3 elements of set B be b-D, b, b+D. Their sum is 3b = 36 implies b = 12. Their product is q = b(b^2 - D^2) = 12(144 - D^2). ### Step 1: Using the Ratio Condition We are given the relation: fracp + qp - q = frac195 Using componendo and dividendo: fracpq = frac19 + 519 - 5 = frac2414 = frac127 Substitute the expression blocks for p and q: frac12(144 - d^2)12(144 - D^2) = frac127 implies frac144 - d^2144 - D^2 = frac127 7(144 - d^2) = 12(144 - D^2) ### Step 2: Substituting D in terms of d We are given D = d + 3: 7(144 - d^2) = 12big(144 - (d + 3)^2big) 1008 - 7d^2 = 12big(144 - (d^2 + 6d + 9)big) 1008 - 7d^2 = 12big(135 - d^2 - 6dbig) = 1620 - 12d^2 - 72d 5d^2 + 72d - 612 = 0 Solving this quadratic equation: (d - 6)(5d + 102) = 0 Since d > 0, we choose d = 6. This implies D = 6 + 3 = 9. ### Step 3: Finding p - q Now calculate the targeted metric: p - q = 12(144 - d^2) - 12(144 - D^2) = 12(D^2 - d^2) p - q = 12(9^2 - 6^2) = 12(81 - 36) = 12(45) = 540 ### Pattern Recognition For 3-element symmetric AP sequences, choosing terms as x-d, x, x+d ensures the sum isolates the middle term instantly (3x = S). This drastically drops algebraic variables from the start. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q53 jee_main_2025_04_april_morning Arithmetic Progression
Let A = \1, 6, 11, 16, dots\ and B = \9, 16, 23, 30, dots\ be the sets consisting of the first 2025 terms of two arithmetic progressions. Then n(A cup B) is
  • A. 3814
  • B. 4027
  • C. 3761
  • D. 4003

Solution

### Related Formula Set Principle of Inclusion-Exclusion: n(A cup B) = n(A) + n(B) - n(A cap B) ### Core Logic Find the last terms of both progressions: For set A: a_1 = 1, d_1 = 5 implies T_2025 = 1 + (2025 - 1) times 5 = 10121. For set B: b_1 = 9, d_2 = 7 implies T_2025 = 9 + (2025 - 1) times 7 = 14177. The intersection set A cap B forms an AP with a common difference d = textLCM(5, 7) = 35. The first common term is 16. ### Step 1: Find Common Terms Count The general term of the common AP must satisfy: T_n = 16 + (n - 1) times 35 le min(10121, 14177) = 10121 (n - 1) times 35 le 10105 implies n - 1 le 288.71 implies n = 289 ### Step 2: Total Distinct Terms Apply the inclusion-exclusion principle: n(A cup B) = 2025 + 2025 - 289 = 3761 ### Pattern Recognition Common terms of two APs always generate a new AP whose common difference is the LCM of the individual common differences. Always verify the upper limit bound using the smaller of the two final values. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequence and Series
Q61 jee_main_2025_04_april_morning Special Series
1 + 3 + 5^2 + 7 + 9^2 + dots upto 40 terms is equal to
  • A. 43890
  • B. 41880
  • C. 33980
  • D. 40870

Solution

### Related Formula Summation Identities: sum r = fracn(n+1)2, quad sum r^2 = fracn(n+1)(2n+1)6 ### Core Logic Split the 40-term series into two sub-series of 20 terms each: Series 1 (squared terms at positions 1, 3, 5... wait, positions are odd numbers whose base squares are odd): 1^2 + 5^2 + 9^2 + dots upto 20 terms. General term T_r = (4r - 3)^2. Series 2 (linear terms at positions 2, 4, 6...): 3 + 7 + 11 + dots upto 20 terms. General term t_r = (4r - 1). ### Step 1: Formulate Total Sigma Expression textSum = sum_r=1^20 left[ (4r - 3)^2 + (4r - 1) right] textSum = sum_r=1^20 (16r^2 - 24r + 9 + 4r - 1) = sum_r=1^20 (16r^2 - 20r + 8) textSum = 16sum_r=1^20 r^2 - 20sum_r=1^20 r + 8sum_r=1^20 1 ### Step 2: Arithmetic Evaluation sum_r=1^20 r^2 = frac20 times 21 times 416 = 2870 sum_r=1^20 r = frac20 times 212 = 210 textSum = 16(2870) - 20(210) + 8(20) = 45920 - 4200 + 160 = 41880 ### Pattern Recognition When dealing with interlaced series, pairing terms adjacent to each other simplifies the degree of general expressions into manageable standard summation polynomials. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequence and Series
Q60 jee_main_2025_07_april_evening Arithmetic Progression
Let a_n be the n^textth term of an A. P. If S_mathrmn = a_1 + a_2 + a_3 + dots + a_mathrmn = 700, a_6 = 7 and S_7 = 7, then a_n is equal to:
  • A. 56
  • B. 65
  • C. 64
  • D. 70

Solution

### Related Formula Sum of first n terms of an AP is given by: S_n = fracn2[2a + (n-1)d] ### Core Logic Given specifications: 1) a_6 = 7 implies a + 5d = 7 quad dots text(ii) 2) S_7 = 7 implies frac72(2a + 6d) = 7 implies a + 3d = 1 quad dots text(iii) Subtracting (iii) from (ii): 2d = 6 implies d = 3 Substituting d=3 into (iii): a + 3(3) = 1 implies a = -8 ### Step 1: Find n from Sn = 700 Substitute a = -8 and d = 3 into the equation for S_n = 700: 700 = fracn2[2(-8) + (n-1)3] 1400 = n[-16 + 3n - 3] 3n^2 - 19n - 1400 = 0 Factoring the quadratic equation: (3n + 56)(n - 25) = 0 Since n must be a positive integer, n = 25. ### Step 2: Determine standard term value We need to find a_25 corresponding to index n=25: a_25 = a + 24d a_25 = -8 + 24(3) = -8 + 72 = 64 ### Pattern Recognition When S_n and specific terms are given, prioritize finding the first term a and common difference d through simple elimination headers before targeting the value of n. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q70 jee_main_2025_07_april_evening Geometric Progression
If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is :
  • A. 745
  • B. 755
  • C. 750
  • D. 757

Solution

### Related Formula Sum of first n terms of a Geometric Progression (GP) is: S_n = fraca(r^n - 1)r - 1 ### Core Logic Let the first term be a and common ratio be r. Given: 1) ar + ar^3 + ar^5 = 21 implies ar(1 + r^2 + r^4) = 21 quad dots text(1) 2) ar^7 + ar^9 + ar^11 = 15309 implies ar^7(1 + r^2 + r^4) = 15309 quad dots text(2) Dividing equation (2) by equation (1): fracar^7ar = frac1530921 implies r^6 = 729 implies r = 3 ### Step 1: Solve for a Substitute r = 3 into equation (1): a(3)(1 + 9 + 81) = 21 3a(91) = 21 implies a = frac791 = frac113 ### Step 2: Find Sum of 9 terms Evaluating S_9: S_9 = fraca(r^9 - 1)r - 1 = fracfrac113(3^9 - 1)3 - 1 = frac19683 - 126 = frac1968226 = 757 ### Pattern Recognition Ratios of shifted groups of terms in a GP always cleanly isolate a simple power of the common ratio r^k instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

More Sequences and Series Questions — jee_main_2026_21_jan_morning

Practice all Sequences and Series previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)
YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Now Live Open
PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics

Exam Analysis Tools

Chapter weightage trends and topic-wise practice — all mapped from past shift papers.

Year-wise PYQ

Chapter Weightage & Trends

See high-yield chapters, year-wise counts and the frequency-based chance each topic repeats next exam.

Weightage Board Open
Targeted PYQs

Topic-wise Practice Hub

Every past-year question for a chapter pulled from all papers, with its key formulas alongside.

One chapter at a time Open
Latest from the Blog
View all →

Loading articles...