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$11^{\text{th}}$ term is:

### 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