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

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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

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