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:

### Related Formula Sum of first n$n$ terms of an Arithmetic Progression: S_n = fracn2[2a + (n-1)d]$$S_n = \frac{n}{2}[2a + (n-1)d]$$ ### Core Logic Set up linear expressions for the given sums : S_40 = frac402[2a + 39d] = 1030 Rightarrow 2a + 39d = 51.5$$S_{40} = \frac{40}{2}[2a + 39d] = 1030 \Rightarrow 2a + 39d = 51.5$$ S_12 = frac122[2a + 11d] = 57 Rightarrow 2a + 11d = 9.5$$S_{12} = \frac{12}{2}[2a + 11d] = 57 \Rightarrow 2a + 11d = 9.5$$ ### Step 1: Solve for a$a$ and d$d$ Subtract the second equation from the first : (2a + 39d) - (2a + 11d) = 51.5 - 9.5$$(2a + 39d) - (2a + 11d) = 51.5 - 9.5$$ 28d = 42 Rightarrow d = frac4228 = frac32 = 1.5$$28d = 42 \Rightarrow d = \frac{42}{28} = \frac{3}{2} = 1.5$$ Substitute d = 1.5$d = 1.5$ back to find a$a$: 2a + 11(1.5) = 9.5 Rightarrow 2a + 16.5 = 9.5 Rightarrow 2a = -7 Rightarrow a = -3.5$$2a + 11(1.5) = 9.5 \Rightarrow 2a + 16.5 = 9.5 \Rightarrow 2a = -7 \Rightarrow a = -3.5$$ ### Step 2: Evaluate S_30 - S_10$S_{30} - S_{10}$ Write out the formula for the target subtraction : S_30 - S_10 = frac302[2a + 29d] - frac102[2a + 9d]$$S_{30} - S_{10} = \frac{30}{2}[2a + 29d] - \frac{10}{2}[2a + 9d]$$ = 15(2a + 29d) - 5(2a + 9d) = 30a + 435d - 10a - 45d = 20a + 390d$$= 15(2a + 29d) - 5(2a + 9d) = 30a + 435d - 10a - 45d = 20a + 390d$$ [cite: 3964, 3965] Substitute the values of a$a$ and d$d$ : = 20(-3.5) + 390(1.5) = -70 + 585 = 515$$= 20(-3.5) + 390(1.5) = -70 + 585 = 515$$ ### Pattern Recognition Notice that S_30 - S_10$S_{30} - S_{10}$ represents the \sum of terms from T_11$T_{11}$ to T_30$T_{30}$, which can also be formulated as 20 times A_20.5$20 \times A_{20.5}$, saving algebraic steps if calculated symmetrically. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula Sum of an infinite geometric progression: S_infty = fraca1-r quad textfor |r| < 1$$S_{\infty} = \frac{a}{1-r} \quad \text{for } |r| < 1$$ ### Core Logic The given expression is an infinite Arithmetico-Geometric Progression (AGP) : S = 5 + frac5+alpha7 + frac5+2alpha7^2 + frac5+3alpha7^3 + dots infty$$S = 5 + \frac{5+\alpha}{7} + \frac{5+2\alpha}{7^2} + \frac{5+3\alpha}{7^3} + \dots \infty$$ ### Step 1: Shift and Subtract Multiply the equation by the common ratio frac17$\frac{1}{7}$ and shift it by one position : frac17S = frac57 + frac5+alpha7^2 + frac5+2alpha7^3 + dots infty$$\frac{1}{7}S = \frac{5}{7} + \frac{5+\alpha}{7^2} + \frac{5+2\alpha}{7^3} + \dots \infty$$ Subtract this from the original equation: S - frac17S = 5 + left(frac5+alpha-57right) + left(frac5+2alpha-(5+alpha)7^2right) + dots$$S - \frac{1}{7}S = 5 + \left(\frac{5+\alpha-5}{7}\right) + \left(\frac{5+2\alpha-(5+\alpha)}{7^2}\right) + \dots$$ frac67S = 5 + fracalpha7 + fracalpha7^2 + fracalpha7^3 + dots$$\frac{6}{7}S = 5 + \frac{\alpha}{7} + \frac{\alpha}{7^2} + \frac{\alpha}{7^3} + \dots$$ ### Step 2: Sum the Infinite Geometric Series Apply the infinite GP formula to the terms involving alpha$\alpha$ : frac67S = 5 + fracalpha7left(frac11 - frac17right) = 5 + fracalpha7left(frac76right) = 5 + fracalpha6$$\frac{6}{7}S = 5 + \frac{\alpha}{7}\left(\frac{1}{1 - \frac{1}{7}}\right) = 5 + \frac{\alpha}{7}\left(\frac{7}{6}\right) = 5 + \frac{\alpha}{6}$$ Given that S = 7$S = 7$ : frac67(7) = 5 + fracalpha6 Rightarrow 6 = 5 + fracalpha6$$\frac{6}{7}(7) = 5 + \frac{\alpha}{6} \Rightarrow 6 = 5 + \frac{\alpha}{6}$$ 1 = fracalpha6 Rightarrow alpha = 6$$1 = \frac{\alpha}{6} \Rightarrow \alpha = 6$$ ### Pattern Recognition Standard trick for infinite AGPs: Multiply by the common ratio r$r$, shift, and subtract to condense the arithmetic progression component into a straightforward infinite geometric progression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### 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