Let < a_n >$< a_n >$ be a sequence such that a_0 = 0, a_1 = frac12$a_0 = 0, a_1 = \frac{1}{2}$ and 2a_n + 2 = 5a_n + 1 - 3a_n, n = 0, 1, 2, 3, dots$2a_{n + 2} = 5a_{n + 1} - 3a_n, n = 0, 1, 2, 3, \dots$. Then sum_k = 1^100 a_k$\sum_{k = 1}^{100} a_k$ is equal to: (1) 3a_99 - 100$3a_{99} - 100$ (2) 3a_100 - 100$3a_{100} - 100$ (3) 3a_100 + 100$3a_{100} + 100$ (4) 3a_99 + 100$3a_{99} + 100$

### Core Logic Let the 3 elements of set A$A$ in A.P. be a-d, a, a+d$a-d, a, a+d$. Their sum is 3a = 36 implies a = 12$3a = 36 \implies a = 12$. Their product is p = a(a^2 - d^2) = 12(144 - d^2)$p = a(a^2 - d^2) = 12(144 - d^2)$. Similarly, let the 3 elements of set B$B$ be b-D, b, b+D$b-D, b, b+D$. Their sum is 3b = 36 implies b = 12$3b = 36 \implies b = 12$. Their product is q = b(b^2 - D^2) = 12(144 - D^2)$q = b(b^2 - D^2) = 12(144 - D^2)$. ### Step 1: Using the Ratio Condition We are given the relation: fracp + qp - q = frac195$$\frac{p + q}{p - q} = \frac{19}{5}$$ Using componendo and dividendo: fracpq = frac19 + 519 - 5 = frac2414 = frac127$$\frac{p}{q} = \frac{19 + 5}{19 - 5} = \frac{24}{14} = \frac{12}{7}$$ Substitute the expression blocks for p$p$ and q$q$: frac12(144 - d^2)12(144 - D^2) = frac127 implies frac144 - d^2144 - D^2 = frac127$$\frac{12(144 - d^2)}{12(144 - D^2)} = \frac{12}{7} \implies \frac{144 - d^2}{144 - D^2} = \frac{12}{7}$$ 7(144 - d^2) = 12(144 - D^2)$$7(144 - d^2) = 12(144 - D^2)$$ ### Step 2: Substituting D in terms of d We are given D = d + 3$D = d + 3$: 7(144 - d^2) = 12big(144 - (d + 3)^2big)$$7(144 - d^2) = 12\big(144 - (d + 3)^2\big)$$ 1008 - 7d^2 = 12big(144 - (d^2 + 6d + 9)big)$$1008 - 7d^2 = 12\big(144 - (d^2 + 6d + 9)\big)$$ 1008 - 7d^2 = 12big(135 - d^2 - 6dbig) = 1620 - 12d^2 - 72d$$1008 - 7d^2 = 12\big(135 - d^2 - 6d\big) = 1620 - 12d^2 - 72d$$ 5d^2 + 72d - 612 = 0$$5d^2 + 72d - 612 = 0$$ Solving this quadratic equation: (d - 6)(5d + 102) = 0$$(d - 6)(5d + 102) = 0$$ Since d > 0$d > 0$, we choose d = 6$d = 6$. This implies D = 6 + 3 = 9$D = 6 + 3 = 9$. ### 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)$$p - q = 12(144 - d^2) - 12(144 - D^2) = 12(D^2 - d^2)$$ p - q = 12(9^2 - 6^2) = 12(81 - 36) = 12(45) = 540$$p - q = 12(9^2 - 6^2) = 12(81 - 36) = 12(45) = 540$$ ### Pattern Recognition For 3-element symmetric AP sequences, choosing terms as x-d, x, x+d$x-d, x, x+d$ ensures the sum isolates the middle term instantly (3x = S$3x = S$). This drastically drops algebraic variables from the start. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula Set Principle of Inclusion-Exclusion: n(A cup B) = n(A) + n(B) - n(A cap B)$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$ ### Core Logic Find the last terms of both progressions: For set A$A$: a_1 = 1, d_1 = 5 implies T_2025 = 1 + (2025 - 1) times 5 = 10121$a_1 = 1, d_1 = 5 \implies T_{2025} = 1 + (2025 - 1) \times 5 = 10121$. For set B$B$: b_1 = 9, d_2 = 7 implies T_2025 = 9 + (2025 - 1) times 7 = 14177$b_1 = 9, d_2 = 7 \implies T_{2025} = 9 + (2025 - 1) \times 7 = 14177$. The intersection set A cap B$A \cap B$ forms an AP with a common difference d = textLCM(5, 7) = 35$d = \text{LCM}(5, 7) = 35$. The first common term is 16$16$. ### 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$$T_n = 16 + (n - 1) \times 35 \le \min(10121, 14177) = 10121$$ (n - 1) times 35 le 10105 implies n - 1 le 288.71 implies n = 289$$(n - 1) \times 35 \le 10105 \implies n - 1 \le 288.71 \implies n = 289$$ ### Step 2: Total Distinct Terms Apply the inclusion-exclusion principle: n(A cup B) = 2025 + 2025 - 289 = 3761$$n(A \cup B) = 2025 + 2025 - 289 = 3761$$ ### 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

### Related Formula Summation Identities: sum r = fracn(n+1)2, quad sum r^2 = fracn(n+1)(2n+1)6$$\sum r = \frac{n(n+1)}{2}, \quad \sum r^2 = \frac{n(n+1)(2n+1)}{6}$$ ### 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$1^2 + 5^2 + 9^2 + \dots$ upto 20 terms. General term T_r = (4r - 3)^2$T_r = (4r - 3)^2$. Series 2 (linear terms at positions 2, 4, 6...): 3 + 7 + 11 + dots$3 + 7 + 11 + \dots$ upto 20 terms. General term t_r = (4r - 1)$t_r = (4r - 1)$. ### Step 1: Formulate Total Sigma Expression textSum = sum_r=1^20 left[ (4r - 3)^2 + (4r - 1) right]$$\text{Sum} = \sum_{r=1}^{20} \left[ (4r - 3)^2 + (4r - 1) \right]$$ textSum = sum_r=1^20 (16r^2 - 24r + 9 + 4r - 1) = sum_r=1^20 (16r^2 - 20r + 8)$$\text{Sum} = \sum_{r=1}^{20} (16r^2 - 24r + 9 + 4r - 1) = \sum_{r=1}^{20} (16r^2 - 20r + 8)$$ textSum = 16sum_r=1^20 r^2 - 20sum_r=1^20 r + 8sum_r=1^20 1$$\text{Sum} = 16\sum_{r=1}^{20} r^2 - 20\sum_{r=1}^{20} r + 8\sum_{r=1}^{20} 1$$ ### Step 2: Arithmetic Evaluation sum_r=1^20 r^2 = frac20 times 21 times 416 = 2870$$\sum_{r=1}^{20} r^2 = \frac{20 \times 21 \times 41}{6} = 2870$$ sum_r=1^20 r = frac20 times 212 = 210$$\sum_{r=1}^{20} r = \frac{20 \times 21}{2} = 210$$ textSum = 16(2870) - 20(210) + 8(20) = 45920 - 4200 + 160 = 41880$$\text{Sum} = 16(2870) - 20(210) + 8(20) = 45920 - 4200 + 160 = 41880$$ ### 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

### Related Formula Sum of first n$n$ terms of an AP is given by: S_n = fracn2[2a + (n-1)d]$$S_n = \frac{n}{2}[2a + (n-1)d]$$ ### Core Logic Given specifications: 1) a_6 = 7 implies a + 5d = 7 quad dots text(ii)$a_6 = 7 \implies a + 5d = 7 \quad \dots \text{(ii)}$ 2) S_7 = 7 implies frac72(2a + 6d) = 7 implies a + 3d = 1 quad dots text(iii)$S_7 = 7 \implies \frac{7}{2}(2a + 6d) = 7 \implies a + 3d = 1 \quad \dots \text{(iii)}$ Subtracting (iii) from (ii): 2d = 6 implies d = 3$$2d = 6 \implies d = 3$$ Substituting d=3$d=3$ into (iii): a + 3(3) = 1 implies a = -8$$a + 3(3) = 1 \implies a = -8$$ ### Step 1: Find n from Sn = 700 Substitute a = -8$a = -8$ and d = 3$d = 3$ into the equation for S_n = 700$S_n = 700$: 700 = fracn2[2(-8) + (n-1)3]$$700 = \frac{n}{2}[2(-8) + (n-1)3]$$ 1400 = n[-16 + 3n - 3]$$1400 = n[-16 + 3n - 3]$$ 3n^2 - 19n - 1400 = 0$$3n^2 - 19n - 1400 = 0$$ Factoring the quadratic equation: (3n + 56)(n - 25) = 0$$(3n + 56)(n - 25) = 0$$ Since n$n$ must be a positive integer, n = 25$n = 25$. ### Step 2: Determine standard term value We need to find a_25$a_{25}$ corresponding to index n=25$n=25$: a_25 = a + 24d$$a_{25} = a + 24d$$ a_25 = -8 + 24(3) = -8 + 72 = 64$$a_{25} = -8 + 24(3) = -8 + 72 = 64$$ ### Pattern Recognition When S_n$S_n$ and specific terms are given, prioritize finding the first term a$a$ and common difference d$d$ through simple elimination headers before targeting the value of n$n$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula Sum of first n$n$ terms of a Geometric Progression (GP) is: S_n = fraca(r^n - 1)r - 1$$S_n = \frac{a(r^n - 1)}{r - 1}$$ ### Core Logic Let the first term be a$a$ and common ratio be r$r$. Given: 1) ar + ar^3 + ar^5 = 21 implies ar(1 + r^2 + r^4) = 21 quad dots text(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)$ar^7 + ar^9 + ar^{11} = 15309 \implies ar^7(1 + r^2 + r^4) = 15309 \quad \dots \text{(2)}$ Dividing equation (2) by equation (1): fracar^7ar = frac1530921 implies r^6 = 729 implies r = 3$$\frac{ar^7}{ar} = \frac{15309}{21} \implies r^6 = 729 \implies r = 3$$ ### Step 1: Solve for a Substitute r = 3$r = 3$ into equation (1): a(3)(1 + 9 + 81) = 21$$a(3)(1 + 9 + 81) = 21$$ 3a(91) = 21 implies a = frac791 = frac113$$3a(91) = 21 \implies a = \frac{7}{91} = \frac{1}{13}$$ ### Step 2: Find Sum of 9 terms Evaluating S_9$S_9$: S_9 = fraca(r^9 - 1)r - 1 = fracfrac113(3^9 - 1)3 - 1 = frac19683 - 126 = frac1968226 = 757$$S_9 = \frac{a(r^9 - 1)}{r - 1} = \frac{\frac{1}{13}(3^9 - 1)}{3 - 1} = \frac{19683 - 1}{26} = \frac{19682}{26} = 757$$ ### Pattern Recognition Ratios of shifted groups of terms in a GP always cleanly isolate a simple power of the common ratio r^k$r^k$ instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series