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 Type:
Enter a numerical value Answer: 2 to 2 +4 marks

Solution & Explanation

### 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

Reference Study Guides

Sequences And Series

More Sequences and Series Previous-Year Questions — Page 7

Q6 jee_main_2024_27_jan_morning Arithmetic Progression
The number of common terms in the progressions 4, 9, 14, 19, dots up to 25^th term and 3, 6, 9, 12, dots up to 37^th term is :
  • A. 9
  • B. 5
  • C. 7
  • D. 8

Solution

### Related Formula T_n = a + (n-1)d D_textcommon = textLCM(d_1, d_2) ### Core Logic First Progression (S_1): 4, 9, 14, 19, dots Common difference d_1 = 5. Last term (T_25) = 4 + (25-1)5 = 4 + 120 = 124. Second Progression (S_2): 3, 6, 9, 12, dots Common difference d_2 = 3. Last term (T_37) = 3 + (37-1)3 = 3 + 108 = 111. ### Step 1: Forming the Common AP By inspecting the sequences, the first common term (a_textcommon) is 9. The common difference of the new series is the LCM of the original differences: D_textcommon = textLCM(5, 3) = 15 Thus, the common terms form a new AP: 9, 24, 39, 54, dots ### Step 2: Bounding the Sequence The last term of the common AP must be less than or equal to the smallest maximum limit of the two series. Here, min(124, 111) = 111. So, the n-th term of the common sequence is bounded by 111: 9 + (n-1)15 le 111 15(n-1) le 102 (n-1) le frac10215 = 6.8 n le 7.8 Since n must be an integer, n = 7. ### Pattern Recognition The common terms of two APs always form a new AP. Its common difference is the LCM of the original differences. Find the first common term manually, then cap the n-th term inequality with the smallest end-boundary of the original sets. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series
Q25 jee_main_2024_27_jan_morning Arithmetico-Geometric Progression
If 8 = 3 + frac14(3+p) + frac14^2(3+2p) + frac14^3(3+3p) + dots infty, then the value of p is:
Numerical Answer. Answer: 9 to 9

Solution

### Related Formula S_infty = fraca1-r + fracdr(1-r)^2 (Sum of an infinite Arithmetico-Geometric Progression, where a is the first AP term, d is common difference, and r is geometric ratio). ### Core Logic The series given is an AGP. However, let's look at it explicitly. Let S = 8. 8 = 3 + frac3+p4 + frac3+2p4^2 + dots Multiply the entire equation by the geometric ratio (1/4): frac84 = frac34 + frac3+p4^2 + frac3+2p4^3 + dots ### Step 1: Shift and Subtract Subtract the shifted series from the original series: 8 - frac84 = 3 + left(frac3+p4 - frac34right) + left(frac3+2p4^2 - frac3+p4^2right) + dots 8 - 2 = 3 + fracp4 + fracp4^2 + fracp4^3 + dots 6 = 3 + fracp4 left( 1 + frac14 + frac14^2 + dots right) ### Step 2: Summing the pure Infinite GP The term in parentheses is an infinite geometric series with a=1 and r=1/4. Sum = frac11 - 1/4 = frac13/4 = frac43 ### Step 3: Final Output Evaluation Substitute this sum back: 6 = 3 + fracp4 times frac43 6 - 3 = fracp3 3 = fracp3 Rightarrow p = 9 ### Pattern Recognition The shift-and-subtract technique natively nullifies the arithmetic growth leaving behind a uniform geometric progression. Using the AGP direct formula S = a/(1-r) + dr/(1-r)^2 works perfectly here as well. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series
Q1 jee_main_2024_29_jan_morning Geometric Progression
If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to
  • A. 7
  • B. 4
  • C. 5
  • D. 6

Solution

### Related Formula S_n = fraca(1 - r^n)1 - r where S_n is the sum of n terms, a is the first term, and r is the common ratio. ### Core Logic Let the terms of the G.P. be a, ar, ar^2, ar^3, dots, ar^63. The sum of all 64 terms is given by: S_textall = a + ar + ar^2 + dots + ar^63 = fraca(1 - r^64)1 - r The odd terms are a, ar^2, ar^4, dots, ar^62. This forms another G.P. with 32 terms and a common ratio of r^2. The sum of the odd terms is: S_textodd = a + ar^2 + ar^4 + dots + ar^62 = fraca(1 - (r^2)^32)1 - r^2 = fraca(1 - r^64)1 - r^2 ### Step 1: Equate and Solve for r We are given that S_textall = 7 cdot S_textodd. Substituting our formulas: fraca(1 - r^64)1 - r = 7 cdot fraca(1 - r^64)1 - r^2 Assuming a neq 0 and r neq 1, we can cancel the common terms dots a(1 - r^64) dots from both sides: frac11 - r = frac71 - r^2 Since 1 - r^2 = (1 - r)(1 + r), we have: frac11 - r = frac7(1 - r)(1 + r) 1 + r = 7 r = 6 ### Pattern Recognition Shortcut: In any G.P. with an even number of terms, the ratio of the total sum to the sum of the odd-positioned terms is exactly 1 + r. Thus, 1 + r = 7 Rightarrow r = 6 immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series
Q2 jee_main_2024_29_jan_morning Arithmetic Progression
In an A.P., the sixth term a_6=2. If the product a_1 a_4 a_5 is the greatest, then the common difference of the A.P., is equal to
  • A. frac32
  • B. frac85
  • C. frac23
  • D. frac58

Solution

### Related Formula a_n = a + (n-1)d For finding extrema of a polynomial function f(x), we set its derivative f'(x) = 0. ### Core Logic Given the 6th term of the A.P. is a_6 = 2. a + 5d = 2 Rightarrow a = 2 - 5d We need to maximize the product P = a_1 a_4 a_5. P = a(a + 3d)(a + 4d) Substituting a = 2 - 5d into the expression for P: P = (2 - 5d)(2 - 5d + 3d)(2 - 5d + 4d) P = (2 - 5d)(2 - 2d)(2 - d) ### Step 1: Expand and Differentiate Let's expand P as a function of d, f(d): f(d) = (2 - 5d)(4 - 6d + 2d^2) f(d) = 8 - 12d + 4d^2 - 20d + 30d^2 - 10d^3 f(d) = -10d^3 + 34d^2 - 32d + 8 To find the maximum, we differentiate f(d) with respect to d and equate to zero: f'(d) = -30d^2 + 68d - 32 = 0 15d^2 - 34d + 16 = 0 Factoring the quadratic: 15d^2 - 24d - 10d + 16 = 0 3d(5d - 8) - 2(5d - 8) = 0 (5d - 8)(3d - 2) = 0 This gives critical points d = frac85 and d = frac23. ### Step 2: Check for Maximum We check the second derivative to confirm a maximum: f''(d) = -60d + 68 At d = frac85: f''left(frac85right) = -60left(frac85right) + 68 = -96 + 68 = -28 lt 0 quad (textMaximum) At d = frac23: f''left(frac23right) = -60left(frac23right) + 68 = -40 + 68 = 28 gt 0 quad (textMinimum) Therefore, the greatest product occurs at d = frac85. ### Pattern Recognition When asked to maximize a product of A.P. terms with a known constant term, express all terms strictly in d, build the cubic, and use standard calculus f'(x)=0 checking roots against the 2nd derivative test (Wavy Curve method works beautifully here). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series Class 12 Mathematics: Application of Derivatives
Q6 jee_main_2024_30_january_evening Geometric Progression
Let a and b be two distinct positive real numbers. Let 11^textth term of a GP, whose first term is a and third term is b , is equal to p^textth term of another GP, whose first term is a and fifth term is b . Then p is equal to
  • A. 20
  • B. 25
  • C. 21
  • D. 24

Solution

### Related Formula n^textth text term of a GP: T_n = a r^n-1 ### Core Logic For the first Geometric Progression (GP): First term t_1 = a Third term t_3 = b = a r_1^2 Rightarrow r_1^2 = fracba The 11^textth term is: t_11 = a r_1^10 = a (r_1^2)^5 = a left(fracbaright)^5 For the second Geometric Progression (GP): First term T_1 = a Fifth term T_5 = a r_2^4 = b Rightarrow r_2^4 = fracba Rightarrow r_2 = left(fracbaright)^1/4 ### Step 1: Equating the Terms The p^textth term of the second GP is: T_p = a r_2^p-1 = a left(left(fracbaright)^1/4right)^p-1 = a left(fracbaright)^fracp-14 Given that t_11 = T_p: a left(fracbaright)^5 = a left(fracbaright)^fracp-14 ### Step 2: Solving for p Since a and b are distinct positive real numbers, fracba neq 1. Therefore, we can equate the exponents: 5 = fracp - 14 20 = p - 1 Rightarrow p = 21 ### Pattern Recognition Express the common ratios strictly in terms of powers of (b/a) to bypass isolated radical tracking. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

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