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 Characteristic equation method for standard second-order linear homogeneous recurrence updates: 2x^2 - 5x + 3 = 0$$2x^2 - 5x + 3 = 0$$ ### Core Logic Solving the characteristic equation gives roots x = 1$x = 1$ and x = frac32$x = \frac{3}{2}$. The general solution takes the form: a_n = A(1)^n + Bleft(frac32right)^n$$a_n = A(1)^n + B\left(\frac{3}{2}\right)^n$$ ### Step 1: Evaluating Sequence Parameters Using boundary conditions: For n = 0 implies A + B = 0$n = 0 \implies A + B = 0$ For n = 1 implies A + frac32B = frac12$n = 1 \implies A + \frac{3}{2}B = \frac{1}{2}$ Solving this simple linear system gives B = 1$B = 1$ and A = -1$A = -1$. Thus, the explicit sequence formula is: a_n = -1 + left(frac32right)^n$$a_n = -1 + \left(\frac{3}{2}\right)^n$$ ### Step 2: Summing the Target Range sum_k = 1^100 a_k = sum_k = 1^100 (-1) + sum_k = 1^100 left(frac32right)^k$$\sum_{k = 1}^{100} a_k = \sum_{k = 1}^{100} (-1) + \sum_{k = 1}^{100} \left(\frac{3}{2}\right)^k$$ = -100 + fracfrac32left[left(frac32right)^100 - 1right]frac32 - 1 = -100 + 3left[left(frac32right)^100 - 1right]$$= -100 + \frac{\frac{3}{2}\left[\left(\frac{3}{2}\right)^{100} - 1\right]}{\frac{3}{2} - 1} = -100 + 3\left[\left(\frac{3}{2}\right)^{100} - 1\right]$$ = 3a_100 - 100$$= 3a_{100} - 100$$ ### Pattern Recognition Characteristic roots directly decouple second-order linear loop progressions into basic combinations of clean geometric progressions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

### Related Formula Standard Arithmetic Progression summation template: S_n = fracn2[2a + (n-1)d]$$S_n = \frac{n}{2}[2a + (n-1)d]$$ ### Core Logic Given structural constraints: T_25 = a + 24d = frac120$$T_{25} = a + 24d = \frac{1}{20}$$ 20 cdot frac252left[a + frac120right] = 13 implies a = frac1500$$20 \cdot \frac{25}{2}\left[a + \frac{1}{20}\right] = 13 \implies a = \frac{1}{500}$$ ### Step 1: Finding Parameters and Indices Substituting a = frac1500$a = \frac{1}{500}$ back into a + 24d = frac120$a + 24d = \frac{1}{20}$ gives d = frac1500$d = \frac{1}{500}$. Using the formula for T_m$T_m$: T_m = a + (m-1)d = frac1500 + fracm-1500 = frac125 implies m = 20$$T_m = a + (m-1)d = \frac{1}{500} + \frac{m-1}{500} = \frac{1}{25} \implies m = 20$$ ### Step 2: Computing the Target Segment Sum For m = 20$m = 20$, the target expression becomes: 5(20) sum_r=20^40 T_r = 100 cdot frac212 [T_20 + T_40]$$5(20) \sum_{r=20}^{40} T_r = 100 \cdot \frac{21}{2} [T_{20} + T_{40}]$$ Evaluating the values gives exactly 126. ### Pattern Recognition When a = d$a = d$, the expressions simplify directly to basic multiples of the index position (T_n = n cdot d$T_n = n \cdot d$), cutting down calculation time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

### 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