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 2

Q70 jee_main_2025_03_april_evening Special Series
The sum 1 + frac1+32! + frac1+3+53! + frac1+3+5+74! + dots is equal to
  • A. 6e
  • B. 4e
  • C. 3e
  • D. 2e

Solution

### Related Formula Sum of first r odd natural numbers: sum_k=1^r (2k-1) = r^2 Exponential series expansion: sum_r=0^infty frac1r! = e ### Core Logic Let's find the general r-th term of the series: T_r = frac1 + 3 + 5 + dots + (2r-1)r! = fracr^2r! = fracr(r-1)! ### Step 1: Expressing term in terms of sum limits Let's write r = (r-1) + 1: T_r = fracr-1+1(r-1)! = frac1(r-2)! + frac1(r-1)! Our infinite sum is: S = sum_r=1^infty T_r = sum_r=2^infty frac1(r-2)! + sum_r=1^infty frac1(r-1)! Both sums are standard representations of the exponential expansion. ### Step 2: Summing the parts - First part: sum_r=2^infty frac1(r-2)! = 1 + frac11! + frac12! + dots = e - Second part: sum_r=1^infty frac1(r-1)! = 1 + frac11! + frac12! + dots = e textTotal Sum S = e + e = 2e ### Pattern Recognition The general term containing r^2 in summation with factorials converges to 2e. Remember the shortcut: sum fracr^2r! = 2e, sum fracr^3r! = 5e. It is extremely useful to memorize these common limits. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series Class 12 Mathematics: Limits, Continuity and Differentiability
Q63 jee_main_2025_07_april_morning Arithmetico-Geometric Progression
Let mathbfx_1, mathbfx_2, mathbfx_3, mathbfx_4 be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from mathbfx_1, mathbfx_2, mathbfx_3, mathbfx_4 then the resulting numbers are in an arithmetic progression. Then the value of frac124 (mathbfx_1 mathbfx_2 mathbfx_3 mathbfx_4) is:
  • A. 72
  • B. 18
  • C. 36
  • D. 216

Solution

### Related Formula For a geometric progression, the terms can be set as a, ar, ar^2, ar^3. For three terms A, B, C to be in arithmetic progression, they must satisfy: 2B = A + C ### Core Logic Let the elements be x_1 = a, x_2 = ar, x_3 = ar^2, x_4 = ar^3. After the specified subtractions, the sequence becomes: a - 2, quad ar - 7, quad ar^2 - 9, quad ar^3 - 5 Since this sequence is in AP, we form two separate common difference linear linkages: 2(ar - 7) = (a - 2) + (ar^2 - 9) implies 2ar - 14 = ar^2 + a - 11 implies ar^2 - 2ar + a + 3 = 0 quad dots (1) 2(ar^2 - 9) = (ar - 7) + (ar^3 - 5) implies 2ar^2 - 18 = ar^3 + ar - 12 implies ar^3 - 2ar^2 + ar + 6 = 0 quad dots (2) ### Step 1: Solve the Simultaneous Polynomials Multiply equation (1) by r: ar^3 - 2ar^2 + ar + 3r = 0 quad dots (3) Subtract equation (3) from equation (2): (ar^3 - 2ar^2 + ar + 6) - (ar^3 - 2ar^2 + ar + 3r) = 0 6 - 3r = 0 implies 3r = 6 implies r = 2 Substitute r = 2 back into equation (1): a(2)^2 - 2a(2) + a + 3 = 0 4a - 4a + a + 3 = 0 implies a = -3 ### Step 2: Find the Continuous Product Value The continuous product term is: mathbfx_1mathbfx_2mathbfx_3mathbfx_4 = a cdot ar cdot ar^2 cdot ar^3 = a^4 r^6 mathbfx_1mathbfx_2mathbfx_3mathbfx_4 = (-3)^4 cdot (2)^6 = 81 times 64 = 5184 Now divide by 24 as required: frac124(5184) = 216 ### Pattern Recognition Notice that multiplying the first AP condition equation by r perfectly mimics the structure of the second condition equation except for the absolute scalar value, allowing direct elimination of all polynomial variable indices simultaneously. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q55 jee_main_2025_08_april_evening Infinite Series
If frac11^4 + frac12^4 + frac13^4 + dots infty = fracpi^490, and frac11^4 + frac13^4 + frac15^4 + dots infty = alpha frac12^4 + frac14^4 + frac16^4 + dots infty = beta then fracalphabeta is equal to
  • A. 23
  • B. 18
  • C. 15
  • D. 14

Solution

### Related Formula textTotal Sum = alpha + beta ### Core Logic Factor out common fractions from the even terms component (beta) to represent it as a scalar multiple of the universal sum sequence. ### Step 1: Simplify the Even Terms Series beta = frac12^4 + frac14^4 + frac16^4 + dots = frac12^4 left( frac11^4 + frac12^4 + frac13^4 + dots right) beta = frac116 left( fracpi^490 right) ### Step 2: Express Alpha by Remainder Deduction Since total sum equals alpha + beta: alpha = textTotal Sum - beta = fracpi^490 - frac116 left( fracpi^490 right) = frac1516 left( fracpi^490 right) ### Step 3: Compute the Relative Ratio 決fracalphabeta = fracfrac1516 left( fracpi^490 right)frac116 left( fracpi^490 right) = 15 ### Pattern Recognition For alternating p-series powers like \sum n^{-p}, the even component fractions always condense via factor steps to 2^{-p} \cdot S_{\text{total}}$, decoupling power values cleanly from final simple scalar quotients. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q57 jee_main_2025_29_jan_evening Infinite Geometric Progression
Let mathbfS = mathbfN cup \0\. Define a relation \mathbf{R} from mathbfS to mathbfR by: mathbfR = left\left(x, yright): log_e y = x log_e left(frac25right), x in S, y in R right\. Then, the sum of all the elements in the range of mathbfR is equal to
  • A. frac32
  • B. frac53
  • C. frac109
  • D. frac52

Solution

### Related Formula Sum of an infinite geometric progression with |r| < 1: S_infty = fraca1 - r ### Core Logic From the definition of the relation: log_e y = x log_eleft(frac25 ight) implies log_e y = log_eleft(frac25right)^x implies y = left(frac25 ight)^x
Infinite Geometric Progression diagram for Q57 - JEE Main 2025 Evening
Infinite Geometric Progression diagram for Q57 - JEE Main 2025 Evening
Since x in S = \0, 1, 2, 3, dots\, the output values of y represent elements of the range. ### Step 1: Compute Infinite Sum Generating elements by plugging in values of x: For x = 0 implies y = 1 For x = 1 implies y = frac25 For x = 2 implies y = left(frac25 ight)^2 Sum of elements in the range: textSum = 1 + left(frac25 ight)^1 + left(frac25 ight)^2 + dots = frac11 - frac25 = frac53 ### Pattern Recognition Convert log equations into standard exponential equations right away. A variable index belonging to whole numbers indicates an infinite GP summation scenario. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series Class 11 Mathematics: Relations and Functions
Q73 jee_main_2025_29_jan_evening Arithmetic Progression Properties
Let a_1, a_2, ldots, a_2024 be an Arithmetic Progression such that a_1 + (a_5 + a_10 + a_15 + ldots + a_2020) + a_2024 = 2233. Then a_1 + a_2 + a_3 + ldots + a_2024 is equal to
Numerical Answer. Answer: 11132 to 11132

Solution

### Related Formula Symmetry identity rule inside Arithmetic Progressions: a_k + a_n-k+1 = a_1 + a_n ### Core Logic Group matching paired steps equidistant from sequence boundary ends: a_1 + a_2024 = a_5 + a_2020 = a_10 + a_2015 = dots The sequence of inner indices follows an AP tracking loop: 5, 10, 15, dots, 2020 Calculate internal block element count N: 2020 = 5 + (N-1)5 implies 2015 = 5(N-1) implies N - 1 = 403 implies N = 404 text terms ### Step 1: Simplify Expression Equations Since the inner sequence contains 404 terms, they form exactly 202 symmetrical pairs. Adding a_1 and a_2024 introduces one more pair, resulting in 203 identical sum blocks: 203(a_1 + a_2024) = 2233 implies a_1 + a_2024 = frac2233203 = 11 ### Step 2: Evaluate the Total Sum Using the standard AP sum formula: S_2024 = frac20242(a_1 + a_2024) = 1012 times 11 = 11132 ### Pattern Recognition Progressions possess natural positional balance. Grouping symmetrical index pairs (a_k + a_n-k+1) allows factoring out variable steps right away. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

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