### Related Formula The general term for the provided series is: T_k = frac1k(k+1) = frac1k - frac1k+1$T_k = \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$ This sets up a standard telescoping summation sequence. ### Core Logic Express the sum S_2025$S_{2025}$ via telescoping fractions: S_2025 = sum_k=1^2025 left( frac1k - frac1k+1 right) = left(1 - frac12right) + left(frac12 - frac13right) + dots + left(frac12025 - frac12026right)$S_{2025} = \sum_{k=1}^{2025} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{2025} - \frac{1}{2026}\right)$ S_2025 = 1 - frac12026 = frac20252026$S_{2025} = 1 - \frac{1}{2026} = \frac{2025}{2026}$ ### Step 1: Compute the boundary expression value Substitute S_2025$S_{2025}$ into the expression value: sqrt2026 cdot S_2025 = sqrt2026 cdot frac20252026 = sqrt2025 = 45$\sqrt{2026 \cdot S_{2025}} = \sqrt{2026 \cdot \frac{2025}{2026}} = \sqrt{2025} = 45$ ### Step 2: Apply Arithmetic Progression Summation The sum of the first 6 terms of the A.P. with a = -p$a = -p$ and d = p$d = p$ is equal to 45: Sigma_6 = frac62 [2a + (6-1)d] = 45$\Sigma_6 = \frac{6}{2} [2a + (6-1)d] = 45$ 3 [2(-p) + 5p] = 45$3 [2(-p) + 5p] = 45$ 3 [3p] = 45 implies 9p = 45 implies p = 5$3 [3p] = 45 \implies 9p = 45 \implies p = 5$ ### Step 3: Calculate target absolute term difference The absolute difference between the 20^textth$20^{\text{th}}$ and 15^textth$15^{\text{th}}$ terms of any A.P. depends strictly on the common difference: |A_20 - A_15| = |(a + 19p) - (a + 14p)| = 5p$|A_{20} - A_{15}| = |(a + 19p) - (a + 14p)| = 5p$ 5p = 5(5) = 25$5p = 5(5) = 25$ ### Pattern Recognition The series sequence frac12 + frac16 + frac112 + dots$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \dots$ is the well-known telescoping series sum frac1n(n+1)$\sum \frac{1}{n(n+1)}$. Its sum to n$n$ terms is identically given by fracnn+1$\frac{n}{n+1}$ without requiring manual re-derivation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula Telescopic series decomposition via method of differences: frac1(k+2)(k+3) = frac1k+2 - frac1k+3$\frac{1}{(k+2)(k+3)} = \frac{1}{k+2} - \frac{1}{k+3}$ ### Core Logic Given: a_n = fracn^2+5n+64 = frac(n+2)(n+3)4$a_n = \frac{n^2+5n+6}{4} = \frac{(n+2)(n+3)}{4}$ Therefore, the reciprocal term is: frac1a_k = frac4(k+2)(k+3) = 4 left[ frac1k+2 - frac1k+3 right]$\frac{1}{a_k} = \frac{4}{(k+2)(k+3)} = 4 \left[ \frac{1}{k+2} - \frac{1}{k+3} \right]$ ### Step 1: Compute the Partial Sum S_n = sum_k=1^n frac1a_k = 4 sum_k=1^n left( frac1k+2 - frac1k+3 right)$S_n = \sum_{k=1}^{n} \frac{1}{a_k} = 4 \sum_{k=1}^{n} \left( \frac{1}{k+2} - \frac{1}{k+3} \right)$ Expanding the sum terms: S_n = 4 left[ left(frac13 - frac14right) + left(frac14 - frac15right) + dots + left(frac1n+2 - frac1n+3right) right]$S_n = 4 \left[ \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \dots + \left(\frac{1}{n+2} - \frac{1}{n+3}\right) \right]$ All intermediate terms cancel out: S_n = 4 left[ frac13 - frac1n+3 right] = 4 left[ fracn+3 - 33(n+3) right] = frac4n3(n+3)$S_n = 4 \left[ \frac{1}{3} - \frac{1}{n+3} \right] = 4 \left[ \frac{n+3 - 3}{3(n+3)} \right] = \frac{4n}{3(n+3)}$ ### Step 2: Calculate for n = 2025 For n = 2025$n = 2025$: S_2025 = frac4 times 20253 times (2025 + 3) = frac4 times 20253 times 2028$S_{2025} = \frac{4 \times 2025}{3 \times (2025 + 3)} = \frac{4 \times 2025}{3 \times 2028}$ We need to find 507 times S_2025$507 \times S_{2025}$: 507 times S_2025 = 507 times frac4 times 20253 times 2028$507 \times S_{2025} = 507 \times \frac{4 \times 2025}{3 \times 2028}$ Notice that 2028 = 4 times 507$2028 = 4 \times 507$: 507 times S_2025 = 507 times frac4 times 20253 times (4 times 507) = frac20253 = 675$507 \times S_{2025} = 507 \times \frac{4 \times 2025}{3 \times (4 \times 507)} = \frac{2025}{3} = 675$ ### Pattern Recognition Always look for arithmetic factor groupings at the end of large number sequence questions in JEE. Here recognizing 2028 = 4 times 507$2028 = 4 \times 507$ avoids large multi-digit multiplication. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula Sum of interior angles of an n$n$-sided polygon: S_n = (n - 2) times 180^circ$S_n = (n - 2) \times 180^\circ$ Sum of an Arithmetic Progression: S_n = fracn2 left[ 2a + (n-1)d right]$S_n = \frac{n}{2} \left[ 2a + (n-1)d \right]$ ### Core Logic The angles form an AP with common difference d = 6^circ$d = 6^\circ$. The largest angle is the last term: T_n = 219^circ$T_n = 219^\circ$. a + (n-1)6 = 219 implies a = 219 - 6n + 6 = 225 - 6n$a + (n-1)6 = 219 \implies a = 219 - 6n + 6 = 225 - 6n$ ### Step 1: Set up the sum equation Equating the two forms for the sum of angles: fracn2 left[ 2a + (n-1)6 right] = (n - 2) times 180$\frac{n}{2} \left[ 2a + (n-1)6 \right] = (n - 2) \times 180$ Substitute a = 225 - 6n$a = 225 - 6n$: fracn2 left[ 2(225 - 6n) + 6n - 6 right] = 180n - 360$\frac{n}{2} \left[ 2(225 - 6n) + 6n - 6 \right] = 180n - 360$ fracn2 left[ 450 - 12n + 6n - 6 right] = 180n - 360$\frac{n}{2} \left[ 450 - 12n + 6n - 6 \right] = 180n - 360$ fracn2 left[ 444 - 6n right] = 180n - 360$\frac{n}{2} \left[ 444 - 6n \right] = 180n - 360$ n(222 - 3n) = 180n - 360$n(222 - 3n) = 180n - 360$ 222n - 3n^2 = 180n - 360$222n - 3n^2 = 180n - 360$ 3n^2 - 42n - 360 = 0$3n^2 - 42n - 360 = 0$ ### Step 2: Solve the Quadratic Equation Divide by 3: n^2 - 14n - 120 = 0$n^2 - 14n - 120 = 0$ (n - 20)(n + 6) = 0$(n - 20)(n + 6) = 0$ Since number of sides n$n$ must be positive, n = 20$n = 20$. ### Pattern Recognition Always remember that any interior angle of a convex polygon must be less than 180^circ$180^\circ$. Let's check the smallest angle for n=20$n=20$: a = 225 - 120 = 105^circ$a = 225 - 120 = 105^\circ$, which is completely valid. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula S_n = fracn2 [2a + (n-1)d]$S_n = \frac{n}{2} [2a + (n-1)d]$ a_n = a + (n-1)d$a_n = a + (n-1)d$ ### Core Logic Given S_3 = 54 implies 3a + 3d = 54 implies a + d = 18$S_3 = 54 \implies 3a + 3d = 54 \implies a + d = 18$. Express S_20$S_{20}$ as: S_20 = frac202[2a + 19d] = 10(2a + 19d)$S_{20} = \frac{20}{2}[2a + 19d] = 10(2a + 19d)$ Substitute a = 18 - d$a = 18 - d$ into the expression: S_20 = 10[2(18 - d) + 19d] = 10(36 + 17d)$S_{20} = 10[2(18 - d) + 19d] = 10(36 + 17d)$ ### Step 1: Formulate Inequality and Constraint Bound Given 1600 < S_20 < 1800$1600 < S_{20} < 1800$: 1600 < 10(36 + 17d) < 1800$1600 < 10(36 + 17d) < 1800$ 160 < 36 + 17d < 180$160 < 36 + 17d < 180$ 124 < 17d < 144$124 < 17d < 144$ frac12417 < d < frac14417 implies 7.29 < d < 8.47$\frac{124}{17} < d < \frac{144}{17} \implies 7.29 < d < 8.47$ ### Step 2: Isolate Integer Term parameters Since the sequence consists of positive integers, common difference d$d$ must be an integer implies d = 8$\implies d = 8$. Then a = 18 - 8 = 10$a = 18 - 8 = 10$. ### Step 3: Calculate the 11th Term a_11 = a + 10d = 10 + 10(8) = 90$a_{11} = a + 10d = 10 + 10(8) = 90$ ### Pattern Recognition Diophantine properties (integer conditions) drastically restrict valid inequality windows. Always check parameters for strict divisibility to skip unnecessary computation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula - An Arithmetic Progression (A.P.) with common difference d$d$ sets consecutive terms as: T_n = T_1 + (n-1)d$T_n = T_1 + (n-1)d$ - A Geometric Progression (G.P.) ensures: T_2^2 = T_1 cdot T_3$T_2^2 = T_1 \cdot T_3$ ### Core Logic Since 3, a, b, c$3, a, b, c$ are elements of an A.P., let d$d$ denote the common difference: - a = 3 + d$a = 3 + d$ - b = 3 + 2d$b = 3 + 2d$ - c = 3 + 3d$c = 3 + 3d$ Substituting these values into the sequence configurations of the given G.P. (3, a-1, b+1, c+9$3, a-1, b+1, c+9$): textG.P. terms: 3, \, (3+d-1), \, (3+2d+1), \, (3+3d+9)$\text{G.P. terms: } 3, \, (3+d-1), \, (3+2d+1), \, (3+3d+9)$ textG.P. terms: 3, \, 2+d, \, 4+2d, \, 12+3d$\text{G.P. terms: } 3, \, 2+d, \, 4+2d, \, 12+3d$ ### Step 1: Compute the Common Difference Using the geometric mean property for the first three terms (3, 2+d, 4+2d$3, 2+d, 4+2d$): (2+d)^2 = 3(4 + 2d)$(2+d)^2 = 3(4 + 2d)$ 4 + 4d + d^2 = 12 + 6d$4 + 4d + d^2 = 12 + 6d$ d^2 - 2d - 8 = 0$d^2 - 2d - 8 = 0$ (d-4)(d+2) = 0 implies d = 4 quad textor quad d = -2$(d-4)(d+2) = 0 \implies d = 4 \quad \text{or} \quad d = -2$ ### Step 2: Evaluate both cases for the Progressions - **Case A: If d = 4$d = 4$** The G.P. sequence reads: 3, 6, 12, 24$3, 6, 12, 24$ (common ratio r=2$r=2$, valid layout). The values are: a = 7, b = 11, c = 15$a = 7, b = 11, c = 15$. - **Case B: If d = -2$d = -2$** The G.P. sequence reads: 3, 0, 0, 6$3, 0, 0, 6$ (contains zeros, violating standard geometric definitions). Hence, select d = 4$d = 4$. ### Step 3: Calculate the Final Arithmetic Mean The required arithmetic mean of a, b, c$a, b, c$ is: textArithmetic Mean = fraca+b+c3 = frac7+11+153 = frac333 = 11$\text{Arithmetic Mean} = \frac{a+b+c}{3} = \frac{7+11+15}{3} = \frac{33}{3} = 11$ ### Pattern Recognition Sees: Transition parameters mapping from A.P. linear spacing into G.P. ratios. Shortcut: Notice that the arithmetic mean of a, b, c$a, b, c$ is exactly equal to the middle value b$b$ for any linear sequence. Thus, finding b = 3 + 2(4) = 11$b = 3 + 2(4) = 11$ directly yields the final answer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series