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 6

Q63 jee_main_2025_29_jan_morning Arithmetic Progression
Consider an A.P. of positive integers, whose \sum of the first three terms is 54 and the \sum of the first twenty terms lies between 1600 and 1800. Then its 11^textth term is:
  • A. 84
  • B. 122
  • C. 90
  • D. 108

Solution

### Related Formula S_n = fracn2 [2a + (n-1)d] a_n = a + (n-1)d ### Core Logic Given S_3 = 54 implies 3a + 3d = 54 implies a + d = 18. Express S_20 as: S_20 = frac202[2a + 19d] = 10(2a + 19d) Substitute a = 18 - d into the expression: S_20 = 10[2(18 - d) + 19d] = 10(36 + 17d) ### Step 1: Formulate Inequality and Constraint Bound Given 1600 < S_20 < 1800: 1600 < 10(36 + 17d) < 1800 160 < 36 + 17d < 180 124 < 17d < 144 frac12417 < d < frac14417 implies 7.29 < d < 8.47 ### Step 2: Isolate Integer Term parameters Since the sequence consists of positive integers, common difference d must be an integer implies d = 8. Then a = 18 - 8 = 10. ### Step 3: Calculate the 11th Term a_11 = a + 10d = 10 + 10(8) = 90 ### Pattern Recognition Diophantine properties (integer conditions) drastically restrict valid inequality windows. Always check parameters for strict divisibility to skip unnecessary computation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q17 jee_main_2024_01_february_morning Arithmetic and Geometric Progressions
Let 3, a, b, c be in A.P. and 3, a-1, b+1, c+9 be in G.P. Then, the arithmetic mean of a, b and c is:
  • A. -4
  • B. -1
  • C. 13
  • D. 11

Solution

### Related Formula - An Arithmetic Progression (A.P.) with common difference d sets consecutive terms as: T_n = T_1 + (n-1)d - A Geometric Progression (G.P.) ensures: T_2^2 = T_1 cdot T_3 ### Core Logic Since 3, a, b, c are elements of an A.P., let d denote the common difference: - a = 3 + d - b = 3 + 2d - c = 3 + 3d Substituting these values into the sequence configurations of the given G.P. (3, a-1, b+1, c+9): textG.P. terms: 3, \, (3+d-1), \, (3+2d+1), \, (3+3d+9) textG.P. terms: 3, \, 2+d, \, 4+2d, \, 12+3d ### Step 1: Compute the Common Difference Using the geometric mean property for the first three terms (3, 2+d, 4+2d): (2+d)^2 = 3(4 + 2d) 4 + 4d + d^2 = 12 + 6d d^2 - 2d - 8 = 0 (d-4)(d+2) = 0 implies d = 4 quad textor quad d = -2 ### Step 2: Evaluate both cases for the Progressions - **Case A: If d = 4** The G.P. sequence reads: 3, 6, 12, 24 (common ratio r=2, valid layout). The values are: a = 7, b = 11, c = 15. - **Case B: If d = -2** The G.P. sequence reads: 3, 0, 0, 6 (contains zeros, violating standard geometric definitions). Hence, select d = 4. ### Step 3: Calculate the Final Arithmetic Mean The required arithmetic mean of a, b, c is: textArithmetic Mean = fraca+b+c3 = frac7+11+153 = frac333 = 11 ### Pattern Recognition Sees: Transition parameters mapping from A.P. linear spacing into G.P. ratios. Shortcut: Notice that the arithmetic mean of a, b, c is exactly equal to the middle value b for any linear sequence. Thus, finding b = 3 + 2(4) = 11 directly yields the final answer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q24 jee_main_2024_01_february_morning Common Terms of Two APs
Let 3, 7, 11, 15, ...., 403 and 2, 5, 8, 11,..., 404 be two arithmetic progressions. Then the sum of the common terms in them is equal to
Numerical Answer. Answer: 6699 to 6699

Solution

### Related Formula - General term of an AP: T_n = a + (n-1)d - Sum of n terms of an AP: S_n = fracn2[2a + (n-1)d] - The common difference of a series of common terms between two APs is given by the least common multiple of their respective common differences: d_textcommon = textLCM(d_1, d_2) ### Core Logic Let's analyze both arithmetic progressions: - AP 1: 3, 7, 11, 15, dots, 403 implies First term a_1 = 3, common difference d_1 = 4. - AP 2: 2, 5, 8, 11, dots, 404 implies First term a_2 = 2, common difference d_2 = 3. By observation, the first identical value appearing in both series is 11. Therefore, the new common AP has: - First term a = 11 - Common difference d = textLCM(4, 3) = 12 ### Step 1: Determine the Number of Common Terms The last term T_n of the common AP cannot exceed the boundary upper limits of either individual series (i.e., le 403): T_n = 11 + (n-1)12 le 403 12(n-1) le 392 n-1 le 32.66 implies n = 33 ### Step 2: Calculate the Series Sum Using the AP summation formula for 33 terms: S_33 = frac332 left[ 2(11) + (33-1)12 right] S_33 = frac332 left[ 22 + 32 times 12 right] S_33 = frac332 left[ 22 + 384 right] S_33 = frac332 times 406 = 33 times 203 = 6699 ### Pattern Recognition Sees: Overlapping arithmetic progression series elements. Shortcut: Once the first matching number and the LCM are calculated, the maximum term inequality a + (n-1)d le min(L_1, L_2) directly maps out the total number of terms cleanly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q9 jee_main_2024_29_january_evening Arithmetic Progression
If log_mathrmea, log_mathrmeb, log_mathrmec are in an A.P. and log_mathrmea - log_mathrme2b, log_mathrme2b - log_mathrme3c, log_mathrme3c - log_mathrmea are also in an A.P, then a:b:c is equal to
  • A. 9 : 6 : 4
  • B. 16 : 4 : 1
  • C. 25 : 10 : 4
  • D. 6 : 3 : 2

Solution

### Related Formula If x, y, z are in A.P., then 2y = x + z. ### Core Logic From the first sequence condition: 2log_e b = log_e a + log_e c implies log_e b^2 = log_e(ac) implies b^2 = ac quad dots (i) From the second sequence condition, the components are log_eleft(fraca2bright), log_eleft(frac2b3cright), log_eleft(frac3caright): 2log_eleft(frac2b3cright) = log_eleft(fraca2bright) + log_eleft(frac3caright) left(frac2b3cright)^2 = fraca2b times frac3ca = frac3c2b frac4b^29c^2 = frac3c2b implies 8b^3 = 27c^3 implies fracbc = frac32 quad dots (ii) ### Step 1: Finding Ratios Substituting c = frac2b3 into equation (i): b^2 = a left(frac2b3right) implies b = frac2a3 implies fracab = frac32 Thus, consolidating all parts: a : b = 9 : 6 b : c = 6 : 4 a : b : c = 9 : 6 : 4 ### Pattern Recognition Logarithmic A.P. strings immediately translate to simple geometric proportions inside the core arguments via logarithmic properties (2log x = log x^2). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q13 jee_main_2024_29_january_evening Geometric Progression
If each term of a geometric progression a_1, a_2, a_3, ldots with a_1 = frac18 and a_2 neq a_1, is the arithmetic mean of the next two terms and S_n = a_1 + a_2 + ldots + a_n, then S_20 - S_18 is equal to
  • A. 2^15
  • B. -2^18
  • C. 2^18
  • D. -2^15

Solution

### Related Formula 2a_n = a_n+1 + a_n+2 ### Core Logic Let the terms of the Geometric Progression have a common ratio r. Substituting the geometric forms into the arithmetic mean relationship: 2(a r^n-1) = a r^n + a r^n+1 Dividing out non-zero fields a r^n-1: 2 = r + r^2 implies r^2 + r - 2 = 0 (r + 2)(r - 1) = 0 Since a_2 neq a_1, we have r neq 1. Thus, the common ratio is r = -2. ### Step 1: Evaluating the Target Partial Difference We need to evaluate: S_20 - S_18 = T_19 + T_20 T_19 + T_20 = a r^18 + a r^19 = a r^18(1 + r) Substituting a = frac18 and r = -2: T_19 + T_20 = frac18 (-2)^18 (1 - 2) = frac12^3 cdot 2^18 cdot (-1) = -2^15 ### Pattern Recognition Partial sum differences simplify into standard standalone term values (S_n - S_n-2 = T_n + T_n-1). This eliminates the need to apply long fraction sum formats. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

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