Let a_1 , a_2 , a_3 , ..... be a G.P. of increasing positive terms such that a_2 cdot a_3 cdot a_4 = 64 and a_1 + a_3 + a_5 = frac8137 . Then a_3 + a_5 + a_7 is equal to :

Solution & Explanation

### Related Formula ntext-th term of a GP: T_n = a r^n-1 ### Core Logic Let the terms of the GP be a, ar, ar^2, ar^3, dots Given: a_2 cdot a_3 cdot a_4 = 64 (ar) cdot (ar^2) cdot (ar^3) = 64 a^3 r^6 = 64 Rightarrow (ar^2)^3 = 64 Rightarrow ar^2 = 4 Since it is a GP of increasing positive terms, r > 1 and a > 0. ### Step 1: Utilize the sum condition Given: a_1 + a_3 + a_5 = frac8137 a + ar^2 + ar^4 = frac8137 Extract a from ar^2 = 4 Rightarrow a = frac4r^2. Substitute this in: frac4r^2 + 4 + 4r^2 = frac8137 4left(frac1r^2 + 1 + r^2right) = frac8137 Let r^2 = t: 4left(frac1t + 1 + tright) = frac8137 Wait, there is a much faster method by just scaling the required expression. ### Step 2: Calculate the required expression We need a_3 + a_5 + a_7 = ar^2 + ar^4 + ar^6. Notice that ar^2 + ar^4 + ar^6 = r^2 (a + ar^2 + ar^4). So, required sum = r^2 left(frac8137right). To find r^2, we solve the quadratic in t = r^2: 4left(fract^2 + t + 1tright) = frac8137 28t^2 + 28t + 28 = 813t 28t^2 - 785t + 28 = 0 The roots are t = 28 and t = frac128. Since the GP is increasing, r > 1 Rightarrow r^2 = 28. ### Step 3: Final evaluation Alternatively, expand directly: ar^2 (1 + r^2 + r^4) = 4 (1 + 28 + (28)^2) = 4(1 + 28 + 784) = 4(813) = 3252. (Note: r^2 times frac8137 = 28 times frac8137 = 4 times 813 = 3252) ### Pattern Recognition In GP questions demanding a sum shifted by a fixed index (like a_1+a_3+a_5 to a_3+a_5+a_7), immediately look to factor out the common ratio multiplier r^k. Here it's a simple scaling by r^2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series Class 10 Maths: Quadratic Equations

Reference Study Guides

Sequences And Series

More Sequences and Series Previous-Year Questions — Page 3

Q59 jee_main_2025_28_jan_morning Recurrence Relations and Summation
Let < a_n > be a sequence such that a_0 = 0, a_1 = frac12 and 2a_n + 2 = 5a_n + 1 - 3a_n, n = 0, 1, 2, 3, dots. Then sum_k = 1^100 a_k is equal to: (1) 3a_99 - 100 (2) 3a_100 - 100 (3) 3a_100 + 100 (4) 3a_99 + 100
  • A. 3a_99 - 100
  • B. 3a_100 - 100
  • C. 3a_100 + 100
  • D. 3a_99 + 100

Solution

### Related Formula Characteristic equation method for standard second-order linear homogeneous recurrence updates: 2x^2 - 5x + 3 = 0 ### Core Logic Solving the characteristic equation gives roots x = 1 and x = frac32. The general solution takes the form: a_n = A(1)^n + Bleft(frac32right)^n ### Step 1: Evaluating Sequence Parameters Using boundary conditions: For n = 0 implies A + B = 0 For n = 1 implies A + frac32B = frac12 Solving this simple linear system gives B = 1 and A = -1. Thus, the explicit sequence formula is: a_n = -1 + left(frac32right)^n ### Step 2: Summing the Target Range sum_k = 1^100 a_k = sum_k = 1^100 (-1) + sum_k = 1^100 left(frac32right)^k = -100 + fracfrac32left[left(frac32right)^100 - 1right]frac32 - 1 = -100 + 3left[left(frac32right)^100 - 1right] = 3a_100 - 100 ### Pattern Recognition Characteristic roots directly decouple second-order linear loop progressions into basic combinations of clean geometric progressions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series
Q61 jee_main_2025_28_jan_morning Arithmetic Progression Properties
Let T_r be the r^textth term of an A.P. If for some m, T_m = frac125, T_25 = frac120 and 20sum_r = 1^25 T_r = 13 then 5msum_r = m^2m T_r is equal to: (1) 112 (2) 126 (3) 98 (4) 142
  • A. 112
  • B. 126
  • C. 98
  • D. 142

Solution

### Related Formula Standard Arithmetic Progression summation template: S_n = fracn2[2a + (n-1)d] ### Core Logic Given structural constraints: T_25 = a + 24d = frac120 20 cdot frac252left[a + frac120right] = 13 implies a = frac1500 ### Step 1: Finding Parameters and Indices Substituting a = frac1500 back into a + 24d = frac120 gives d = frac1500. Using the formula for T_m: T_m = a + (m-1)d = frac1500 + fracm-1500 = frac125 implies m = 20 ### Step 2: Computing the Target Segment Sum For m = 20, the target expression becomes: 5(20) sum_r=20^40 T_r = 100 cdot frac212 [T_20 + T_40] Evaluating the values gives exactly 126. ### Pattern Recognition When a = d, the expressions simplify directly to basic multiples of the index position (T_n = n cdot d), cutting down calculation time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series
Q57 jee_main_2025_03_april_morning Method of Differences
The sum 1 + 3 + 11 + 25 + 45 + 71 + dots up to 20 terms, is equal to[cite: 580, 584]:
  • A. 7240
  • B. 7130
  • C. 6982
  • D. 8124

Solution

### Related Formula For series whose consecutive differences form an Arithmetic Progression (A.P.), the general term is given by a quadratic form: T_n = an^2 + bn + c ### Core Logic Analyze successive first-order differences of the terms [cite: 1293, 1294]: textSeries: 1, quad 3, quad 11, quad 25, quad 45, quad 71 [cite: 1293] textDifferences: 2, quad 8, quad 14, quad 20, quad 26 [cite: 1294] Since the differences grow uniformly by 6, they reside in an A.P. [cite: 1294] Thus, set up the general term system [cite: 1296, 1298]: - T_1 = a + b + c = 1 - T_2 = 4a + 2b + c = 3 - T_3 = 9a + 3b + c = 11 Solving the linear equations simultaneously yields [cite: 1299]: a = 3, quad b = -7, quad c = 5 [cite: 1299] ### Step 1: Summing the Series The general term is [cite: 1300]: T_n = 3n^2 - 7n + 5 [cite: 1300] Evaluate the summation for n=20 terms [cite: 1302]: S_20 = sum_n=1^20 (3n^2 - 7n + 5) = 3sum_n=1^20 n^2 - 7sum_n=1^20 n + sum_n=1^20 5 [cite: 1302] Substitute standard sequence formulas [cite: 1302]: S_20 = 3 cdot left(frac20 cdot 21 cdot 416right) - 7 cdot left(frac20 cdot 212right) + 5(20) [cite: 1302] = 8610 - 1470 + 100 = 7240 [cite: 1302] ### Pattern Recognition When first-order differences form a regular arithmetic line, the original function is exactly quadratic. Identify coefficients using small terms quickly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q66 jee_main_2025_03_april_morning Geometric Progression
Let a_1, a_2, a_3, ldots be a G.P. of increasing positive numbers[cite: 655]. If a_3a_5 = 729 and a_2 + a_4 = frac1114 [cite: 656], then 24(a_1 + a_2 + a_3) is equal to[cite: 657]:
  • A. 131
  • B. 130
  • C. 129
  • D. 128

Solution

### Related Formula For a geometric sequence configuration with first term a and common ratio r: a_n = a cdot r^n-1 ### Core Logic Convert information markers using parameter notations [cite: 1372, 1373, 1376]: a_3 a_5 = (ar^2)(ar^4) = a^2 r^6 = 729 implies ar^3 = 27 [cite: 1373, 1374] From second expression block data [cite: 1376]: a_2 + a_4 = ar + ar^3 = frac1114 [cite: 1376] Substitute ar^3 = 27 directly into the linear equation block [cite: 1376]: ar + 27 = frac1114 implies ar = frac1114 - 27 = frac34 [cite: 1376] ### Step 1: Finding parameters a and r Divide the calculated components to evaluate the ratio [cite: 1387]: fracar^3ar = frac273/4 implies r^2 = 36 implies r = 6 [cite: 1387] (Choose +6 because terms must stay strictly positive [cite: 655]). Find value for first base variable a [cite: 1389]: a(6) = frac34 implies a = frac18 [cite: 1389] ### Step 2: Sum configuration resolving Now compute targeted expansion expression value [cite: 1390]: 24(a_1 + a_2 + a_3) = 24(a + ar + ar^2) = 24a(1 + r + r^2) [cite: 1390] = 24 cdot left(frac18right) cdot (1 + 6 + 36) = 3 cdot 43 = 129 [cite: 1390, 1391] ### Pattern Recognition Product entries like a_3 a_5 = a_4^2 help identify the central term index value quickly in symmetric geometric progressions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q62 jee_main_2025_04_april_evening Telescoping Series
If the sum of the first 20 terms of the series frac 4 . 14 + 3 . 1 ^ 2 + 1 ^ 4 + frac 4 . 24 + 3 . 2 ^ 2 + 2 ^ 4 + frac 4 . 34 + 3 . 3 ^ 2 + 3 ^ 4 + frac 4 . 44 + 3 . 4 ^ 2 + 4 ^ 4 + dots is fracmathrmmmathrmn, where m and n are coprime, then mathrmm + mathrmn is equal to:
  • A. 423
  • B. 420
  • C. 421
  • D. 422

Solution

### Core Logic The general term T_r of the series can be written as: T_r = frac4rr^4 + 3r^2 + 4 Let's factorize the denominator by completing the square metric: r^4 + 3r^2 + 4 = (r^4 + 4r^2 + 4) - r^2 = (r^2 + 2)^2 - r^2 Using the difference of squares identity A^2 - B^2 = (A-B)(A+B): r^4 + 3r^2 + 4 = (r^2 - r + 2)(r^2 + r + 2) ### Step 1: Partial Fraction Decomposition Express T_r using partial fractions split: T_r = frac4r(r^2 - r + 2)(r^2 + r + 2) = 2 left[ frac1r^2 - r + 2 - frac1r^2 + r + 2 right] Notice that if we define V(r) = r^2 - r + 2, then V(r+1) = (r+1)^2 - (r+1) + 2 = r^2 + 2r + 1 - r - 1 + 2 = r^2 + r + 2. Thus, T_r = 2big[V(r) - V(r+1)big], which sets up a clear telescoping sum formulation. ### Step 2: Evaluating the Sum of 20 Terms Summing from r = 1 to 20: S_20 = sum_r=1^20 T_r = 2 sum_r=1^20 left[ frac1r^2 - r + 2 - frac1r^2 + r + 2 right] = 2 left[ left(frac12 - frac14right) + left(frac14 - frac18right) + dots + left(frac120^2 - 20 + 2 - frac120^2 + 20 + 2right) right] All sequential middle terms cancel completely, leaving only first and final values: S_20 = 2 left[ frac12 - frac1422 right] = 1 - frac1211 = frac210211 Since 210 and 211 are coprime, m = 210 and n = 211. ### Step 3: Calculating m + n Combining both values: m + n = 210 + 211 = 421 ### Pattern Recognition The polynomial factorization r^4 + a^2r^2 + b^4 is a frequent pattern in series problems. Always complete the square to break it into a product of quadratic expressions, which naturally yields a telescoping sequence. ### 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...