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

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Q22 jee_main_2026_21_jan_morning Telescoping Sums and Recurrence Relations
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. Answer: 2 to 2

Solution

### 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
Q53 jee_main_2025_02_april_evening Arithmetic Progression
The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by frac212. Then the number of terms which are integers in the A.P. is :
  • A. 4
  • B. 10
  • C. 6
  • D. 8

Solution

### Related Formula textSum of an A.P.: S_k = frack2 left( 2a + (k-1)d right) textGeneral term of an A.P.: a_k = a_1 + (k-1)d ### Core Logic Let the A.P. have n terms (where n is even). The terms are divided into n/2 odd-indexed terms and n/2 even-indexed terms. ### Step 1: Set up the even and odd sums Sum of even terms: a_2 + a_4 + dots + a_n = 30 quad text--- (1) Sum of odd terms: a_1 + a_3 + dots + a_n-1 = 24 quad text--- (2) Subtracting equation (2) from (1): (a_2 - a_1) + (a_4 - a_3) + dots + (a_n - a_n-1) = 30 - 24 = 6 Since there are n/2 such pairs, and the difference of adjacent terms is the common difference d: fracn2 d = 6 implies n d = 12 quad text--- (3) ### Step 2: Solve for n and d We are given that the last term exceeds the first by frac212: a_n - a_1 = (n-1)d = frac212 n d - d = 10.5 Substitute nd = 12 from (3): 12 - d = 10.5 implies d = 1.5 = frac32 Using this in (3): n left(frac32right) = 12 implies n = 8 ### Step 3: Solve for the first term The sum of the odd terms is: S_textodd = frac42 left[ 2a_1 + (4-1)(2d) right] = 24 2 left[ 2a_1 + 3(3) right] = 24 implies 2a_1 + 9 = 12 implies a_1 = 1.5 = frac32 Thus, the terms are: frac32, \, 3, \, frac92, \, 6, \, frac152, \, 9, \, frac212, \, 12 The terms that are integers are 3, 6, 9, 12. The total number of integer terms is 4. ### Pattern Recognition Sum of even terms minus sum of odd terms in any A.P. with an even number of terms n is always equal to fracn2 d. This is an extremely useful relation to remember. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q72 jee_main_2025_02_april_evening Sum of Special Series
If the sum of the first 10 terms of the series frac4 cdot 11 + 4 cdot 1^4 + frac4 cdot 21 + 4 cdot 2^4 + frac4 cdot 31 + 4 cdot 3^4 + dots is fracmathrmmmathrmn, where gcd(mathrmm, mathrmn) = 1, then mathrmm + mathrmn is equal to ____________.
Numerical Answer. Answer: 441 to 441

Solution

### Related Formula textSophie Germain's algebraic factorization: 1 + 4r^4 = (2r^2 + 2r + 1)(2r^2 - 2r + 1) textTelescoping Series representation: T_r = f(r) - f(r+1) ### Core Logic This is a telescoping series sum. We expand the denominator using Sophie Germain's algebraic identity to write the general term as a difference of two consecutive rational expressions. ### Step 1: Write down the general term and factor The general term T_r of the series is: T_r = frac4r1 + 4r^4 Using the factorization of 1+4r^4: T_r = frac4r(2r^2 - 2r + 1)(2r^2 + 2r + 1) Notice that the numerator 4r is the exact difference of the two quadratic factors: (2r^2 + 2r + 1) - (2r^2 - 2r + 1) = 4r ### Step 2: Split the fraction into telescoping terms Rewrite the general term T_r: T_r = frac(2r^2 + 2r + 1) - (2r^2 - 2r + 1)(2r^2 - 2r + 1)(2r^2 + 2r + 1) T_r = frac12r^2 - 2r + 1 - frac12r^2 + 2r + 1 = f(r) - f(r+1) ### Step 3: Expand the sum and solve Sum the first 10 terms: - For r = 1: T_1 = frac11 - frac15 - For r = 2: T_2 = frac15 - frac113 - ... - For r = 10: T_10 = frac1181 - frac1221 All intermediate terms cancel out: S_10 = 1 - frac1221 = frac220221 = fracmn Since 220 and 221 are coprime (their greatest common divisor is 1): m = 220 quad textand quad n = 221 m + n = 220 + 221 = 441 ### Pattern Recognition Sophie Germain identity: The expansion 4r^4 + 1 = (2r^2 - 2r + 1)(2r^2 + 2r + 1) is highly common in telescoping series. Spotting this factorization collapses the sum instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q jee_main_2025_02_april_morning Arithmetic Progression Summation
Let a_1, a_2, a_3, ldots be in an A.P. such that sum_k=1^12 a_2k-1 = -frac725 a_1, a_1 neq 0. If sum_k=1^n a_k = 0, then n is:
  • A. 11
  • B. 10
  • C. 18
  • D. 17

Solution

### Related Formula Sum of an Arithmetic Progression: S_m = fracm2[2a_1 + (m-1)d] ### Core Logic Express the odd terms summation in terms of a_1 and d, establish their linear dependency, and then solve for n where total sum vanishes. ### Step 1: Simplify the Given Summation The summation represents the sum of 12 terms: a_1 + a_3 + a_5 + dots + a_23. This is an A.P. with initial term a_1 and common difference 2d. frac122[2a_1 + 11(2d)] = -frac725a_1 6[2a_1 + 22d] = -frac725a_1 implies 12a_1 + 132d = -frac725a_1 ### Step 2: Relate initial term to common difference Multiply through by 5 to eliminate the fraction: 60a_1 + 660d = -72a_1 implies 132a_1 + 660d = 0 implies a_1 = -5d ### Step 3: Solve for n Set the general sum of n terms to 0: fracn2[2a_1 + (n-1)d] = 0 implies 2a_1 + (n-1)d = 0 Substitute a_1 = -5d: 2(-5d) + (n-1)d = 0 implies -10d + nd - d = 0 implies nd = 11d Since a_1 neq 0, d neq 0, which yields: n = 11 ### Pattern Recognition The condition a_1 = -5d means the sequence starts positive/negative and counts down symmetrically. A sum of n terms equals zero when the middle term or balanced pairs completely wipe each other out, pointing directly to n = 2(5) + 1 = 11. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

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