### Related Formula For series whose consecutive differences form an Arithmetic Progression (A.P.), the general term is given by a quadratic form: T_n = an^2 + bn + c$T_n = an^2 + bn + c$ ### Core Logic Analyze successive first-order differences of the terms [cite: 1293, 1294]: textSeries: 1, quad 3, quad 11, quad 25, quad 45, quad 71$\text{Series: } 1, \quad 3, \quad 11, \quad 25, \quad 45, \quad 71$ [cite: 1293] textDifferences: 2, quad 8, quad 14, quad 20, quad 26$\text{Differences: } 2, \quad 8, \quad 14, \quad 20, \quad 26$ [cite: 1294] Since the differences grow uniformly by 6, they reside in an A.P. [cite: 1294] Thus, set up the general term system [cite: 1296, 1298]: - T_1 = a + b + c = 1$T_1 = a + b + c = 1$ - T_2 = 4a + 2b + c = 3$T_2 = 4a + 2b + c = 3$ - T_3 = 9a + 3b + c = 11$T_3 = 9a + 3b + c = 11$ Solving the linear equations simultaneously yields [cite: 1299]: a = 3, quad b = -7, quad c = 5$a = 3, \quad b = -7, \quad c = 5$ [cite: 1299] ### Step 1: Summing the Series The general term is [cite: 1300]: T_n = 3n^2 - 7n + 5$T_n = 3n^2 - 7n + 5$ [cite: 1300] Evaluate the summation for n=20$n=20$ terms [cite: 1302]: S_20 = sum_n=1^20 (3n^2 - 7n + 5) = 3sum_n=1^20 n^2 - 7sum_n=1^20 n + sum_n=1^20 5$S_{20} = \sum_{n=1}^{20} (3n^2 - 7n + 5) = 3\sum_{n=1}^{20} n^2 - 7\sum_{n=1}^{20} n + \sum_{n=1}^{20} 5$ [cite: 1302] Substitute standard sequence formulas [cite: 1302]: S_20 = 3 cdot left(frac20 cdot 21 cdot 416right) - 7 cdot left(frac20 cdot 212right) + 5(20)$S_{20} = 3 \cdot \left(\frac{20 \cdot 21 \cdot 41}{6}\right) - 7 \cdot \left(\frac{20 \cdot 21}{2}\right) + 5(20)$ [cite: 1302] = 8610 - 1470 + 100 = 7240$= 8610 - 1470 + 100 = 7240$ [cite: 1302] ### Pattern Recognition When first-order differences form a regular arithmetic line, the original function is exactly quadratic. Identify coefficients using small terms quickly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula For a geometric sequence configuration with first term a$a$ and common ratio r$r$: a_n = a cdot r^n-1$a_n = a \cdot r^{n-1}$ ### Core Logic Convert information markers using parameter notations [cite: 1372, 1373, 1376]: a_3 a_5 = (ar^2)(ar^4) = a^2 r^6 = 729 implies ar^3 = 27$a_3 a_5 = (ar^2)(ar^4) = a^2 r^6 = 729 \implies ar^3 = 27$ [cite: 1373, 1374] From second expression block data [cite: 1376]: a_2 + a_4 = ar + ar^3 = frac1114$a_2 + a_4 = ar + ar^3 = \frac{111}{4}$ [cite: 1376] Substitute ar^3 = 27$ar^3 = 27$ directly into the linear equation block [cite: 1376]: ar + 27 = frac1114 implies ar = frac1114 - 27 = frac34$ar + 27 = \frac{111}{4} \implies ar = \frac{111}{4} - 27 = \frac{3}{4}$ [cite: 1376] ### Step 1: Finding parameters a and r Divide the calculated components to evaluate the ratio [cite: 1387]: fracar^3ar = frac273/4 implies r^2 = 36 implies r = 6$\frac{ar^3}{ar} = \frac{27}{3/4} \implies r^2 = 36 \implies r = 6$ [cite: 1387] (Choose +6$+6$ because terms must stay strictly positive [cite: 655]). Find value for first base variable a$a$ [cite: 1389]: a(6) = frac34 implies a = frac18$a(6) = \frac{3}{4} \implies a = \frac{1}{8}$ [cite: 1389] ### Step 2: Sum configuration resolving Now compute targeted expansion expression value [cite: 1390]: 24(a_1 + a_2 + a_3) = 24(a + ar + ar^2) = 24a(1 + r + r^2)$24(a_1 + a_2 + a_3) = 24(a + ar + ar^2) = 24a(1 + r + r^2)$ [cite: 1390] = 24 cdot left(frac18right) cdot (1 + 6 + 36) = 3 cdot 43 = 129$= 24 \cdot \left(\frac{1}{8}\right) \cdot (1 + 6 + 36) = 3 \cdot 43 = 129$ [cite: 1390, 1391] ### Pattern Recognition Product entries like a_3 a_5 = a_4^2$a_3 a_5 = a_4^2$ help identify the central term index value quickly in symmetric geometric progressions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Core Logic The general term T_r$T_r$ of the series can be written as: T_r = frac4rr^4 + 3r^2 + 4$T_r = \frac{4r}{r^4 + 3r^2 + 4}$ Let's factorize the denominator by completing the square metric: r^4 + 3r^2 + 4 = (r^4 + 4r^2 + 4) - r^2 = (r^2 + 2)^2 - r^2$r^4 + 3r^2 + 4 = (r^4 + 4r^2 + 4) - r^2 = (r^2 + 2)^2 - r^2$ Using the difference of squares identity A^2 - B^2 = (A-B)(A+B)$A^2 - B^2 = (A-B)(A+B)$: r^4 + 3r^2 + 4 = (r^2 - r + 2)(r^2 + r + 2)$r^4 + 3r^2 + 4 = (r^2 - r + 2)(r^2 + r + 2)$ ### Step 1: Partial Fraction Decomposition Express T_r$T_r$ using partial fractions split: T_r = frac4r(r^2 - r + 2)(r^2 + r + 2) = 2 left[ frac1r^2 - r + 2 - frac1r^2 + r + 2 right]$T_r = \frac{4r}{(r^2 - r + 2)(r^2 + r + 2)} = 2 \left[ \frac{1}{r^2 - r + 2} - \frac{1}{r^2 + r + 2} \right]$ Notice that if we define V(r) = r^2 - r + 2$V(r) = r^2 - r + 2$, then V(r+1) = (r+1)^2 - (r+1) + 2 = r^2 + 2r + 1 - r - 1 + 2 = r^2 + r + 2$V(r+1) = (r+1)^2 - (r+1) + 2 = r^2 + 2r + 1 - r - 1 + 2 = r^2 + r + 2$. Thus, T_r = 2big[V(r) - V(r+1)big]$T_r = 2\big[V(r) - V(r+1)\big]$, which sets up a clear telescoping sum formulation. ### Step 2: Evaluating the Sum of 20 Terms Summing from r = 1$r = 1$ to 20$20$: S_20 = sum_r=1^20 T_r = 2 sum_r=1^20 left[ frac1r^2 - r + 2 - frac1r^2 + r + 2 right]$S_{20} = \sum_{r=1}^{20} T_r = 2 \sum_{r=1}^{20} \left[ \frac{1}{r^2 - r + 2} - \frac{1}{r^2 + r + 2} \right]$ = 2 left[ left(frac12 - frac14right) + left(frac14 - frac18right) + dots + left(frac120^2 - 20 + 2 - frac120^2 + 20 + 2right) right]$= 2 \left[ \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{8}\right) + \dots + \left(\frac{1}{20^2 - 20 + 2} - \frac{1}{20^2 + 20 + 2}\right) \right]$ All sequential middle terms cancel completely, leaving only first and final values: S_20 = 2 left[ frac12 - frac1422 right] = 1 - frac1211 = frac210211$S_{20} = 2 \left[ \frac{1}{2} - \frac{1}{422} \right] = 1 - \frac{1}{211} = \frac{210}{211}$ Since 210$210$ and 211$211$ are coprime, m = 210$m = 210$ and n = 211$n = 211$. ### Step 3: Calculating m + n Combining both values: m + n = 210 + 211 = 421$m + n = 210 + 211 = 421$ ### Pattern Recognition The polynomial factorization r^4 + a^2r^2 + b^4$r^4 + a^2r^2 + b^4$ is a frequent pattern in series problems. Always complete the square to break it into a product of quadratic expressions, which naturally yields a telescoping sequence. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### 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