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 textSum of AP: S_k = frack2 [2a + (k - 1)d]$$\text{Sum of AP: } S_k = \frac{k}{2} [2a + (k - 1)d]$$ sum_k=1^n k^2 = fracn(n+1)(2n+1)6$$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$$ sum_k=1^n k = fracn(n+1)2$$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$$ ### Core Logic For the arithmetic progression 3, 7, 11, dots$3, 7, 11, \dots$ First term a = 3$a = 3$, Common difference d = 4$d = 4$. The sum of the first k$k$ terms is: S_k = frack2 (2(3) + (k - 1)4) = frack2 (6 + 4k - 4) = frack2 (4k + 2) = 2k^2 + k$$S_k = \frac{k}{2} (2(3) + (k - 1)4) = \frac{k}{2} (6 + 4k - 4) = \frac{k}{2} (4k + 2) = 2k^2 + k$$ ### Step 1: Finding the Sum of Sums Now compute the sum sum_k=1^n S_k$\sum_{k=1}^{n} S_k$: sum_k=1^n S_k = sum_k=1^n (2k^2 + k) = 2sum_k=1^n k^2 + sum_k=1^n k$$\sum_{k=1}^{n} S_k = \sum_{k=1}^{n} (2k^2 + k) = 2\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$$ = 2 left( fracn(n+1)(2n+1)6 right) + fracn(n+1)2$$= 2 \left( \frac{n(n+1)(2n+1)}{6} \right) + \frac{n(n+1)}{2}$$ = n(n+1) left[ frac2(2n+1)6 + frac12 right] = n(n+1) left[ frac2n+13 + frac12 right]$$= n(n+1) \left[ \frac{2(2n+1)}{6} + \frac{1}{2} \right] = n(n+1) \left[ \frac{2n+1}{3} + \frac{1}{2} \right]$$ = n(n+1) left[ frac4n + 2 + 36 right] = fracn(n+1)(4n + 5)6$$= n(n+1) \left[ \frac{4n + 2 + 3}{6} \right] = \frac{n(n+1)(4n + 5)}{6}$$ ### Step 2: Resolving the Inequality Substitute this sum into the given expression: frac6n(n+1) sum_k=1^n S_k = frac6n(n+1) cdot fracn(n+1)(4n+5)6 = 4n + 5$$\frac{6}{n(n+1)} \sum_{k=1}^{n} S_k = \frac{6}{n(n+1)} \cdot \frac{n(n+1)(4n+5)}{6} = 4n + 5$$ We are given the bounds: 40 lt 4n + 5 lt 42$$40 \lt 4n + 5 \lt 42$$ 35 lt 4n lt 37$$35 \lt 4n \lt 37$$ 8.75 lt n lt 9.25$$8.75 \lt n \lt 9.25$$ Since n$n$ must be an integer (representing the number of terms), the only valid integer is n = 9$n = 9$. ### Pattern Recognition Evaluating a 'sum of sums' for an AP effectively requires applying the Sigma k^2$\Sigma k^2$ and Sigma k$\Sigma k$ standard formulas to the generic S_n$S_n$ quadratic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

### Related Formula S_n = fracn2[2a + (n-1)d]$$S_n = \frac{n}{2}[2a + (n-1)d]$$ ### Core Logic Using the sum formula for an AP: S_20 = frac202[2a + 19d] = 790$$S_{20} = \frac{20}{2}[2a + 19d] = 790$$ 10[2a + 19d] = 790$$10[2a + 19d] = 790$$ 2a + 19d = 79 quad dots (1)$$2a + 19d = 79 \quad \dots (1)$$ S_10 = frac102[2a + 9d] = 145$$S_{10} = \frac{10}{2}[2a + 9d] = 145$$ 5[2a + 9d] = 145$$5[2a + 9d] = 145$$ 2a + 9d = 29 quad dots (2)$$2a + 9d = 29 \quad \dots (2)$$ ### Step 1: Solving for a and d Subtracting (2) from (1): 10d = 50 Rightarrow d = 5$$10d = 50 \Rightarrow d = 5$$ Substituting d=5$d=5$ into (2): 2a + 9(5) = 29 Rightarrow 2a = 29 - 45 = -16$$2a + 9(5) = 29 \Rightarrow 2a = 29 - 45 = -16$$ a = -8$a = -8$ ### Step 2: Evaluating the required expression We need to find S_15 - S_5$S_{15} - S_{5}$: S_15 - S_5 = frac152[2a + 14d] - frac52[2a + 4d]$$S_{15} - S_5 = \frac{15}{2}[2a + 14d] - \frac{5}{2}[2a + 4d]$$ Substituting 2a = -16$2a = -16$ and d = 5$d = 5$: = frac152[-16 + 70] - frac52[-16 + 20]$$= \frac{15}{2}[-16 + 70] - \frac{5}{2}[-16 + 20]$$ = frac152[54] - frac52[4]$$= \frac{15}{2}[54] - \frac{5}{2}[4]$$ = 15 times 27 - 5 times 2$$= 15 \times 27 - 5 \times 2$$ = 405 - 10 = 395$$= 405 - 10 = 395$$ ### Pattern Recognition When two sums of an AP are given, immediately set up the linear equations in terms of a$a$ and d$d$. Solve for them, and substitute directly into the target expression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

### Related Formula textMethod of differences for a sequence: V_n - V_n-1 = T_n$$\text{Method of differences for a sequence: } V_n - V_{n-1} = T_n$$ ### Core Logic The base terms inside the squares form a sequence: 1, 4, 8, 13, 19, 26 dots$1, 4, 8, 13, 19, 26 \dots$ The differences between consecutive terms are: 3, 4, 5, 6, 7 dots$3, 4, 5, 6, 7 \dots$ Since the first differences are an Arithmetic Progression, the general term of the inner sequence is a quadratic in n$n$: T_n = an^2 + bn + c$T_n = an^2 + bn + c$. Using n=1$n=1$: 1 = a + b + c$1 = a + b + c$ Using n=2$n=2$: 4 = 4a + 2b + c$4 = 4a + 2b + c$ Using n=3$n=3$: 8 = 9a + 3b + c$8 = 9a + 3b + c$ Solving this system: (4a + 2b + c) - (a + b + c) = 3 Rightarrow 3a + b = 3$(4a + 2b + c) - (a + b + c) = 3 \Rightarrow 3a + b = 3$ (9a + 3b + c) - (4a + 2b + c) = 4 Rightarrow 5a + b = 4$(9a + 3b + c) - (4a + 2b + c) = 4 \Rightarrow 5a + b = 4$ Subtracting these gives: 2a = 1 Rightarrow a = 1/2$2a = 1 \Rightarrow a = 1/2$. Then 3(1/2) + b = 3 Rightarrow b = 3/2$3(1/2) + b = 3 \Rightarrow b = 3/2$. Finally 1/2 + 3/2 + c = 1 Rightarrow c = -1$1/2 + 3/2 + c = 1 \Rightarrow c = -1$. Inner sequence T_n = frac12n^2 + frac32n - 1$T_n = \frac{1}{2}n^2 + \frac{3}{2}n - 1$. ### Step 1: Calculating alpha structure The series is alpha = sum_n=1^10 (T_n)^2$\alpha = \sum_{n=1}^{10} (T_n)^2$. 4alpha = sum_n=1^10 4left(fracn^2 + 3n - 22right)^2 = sum_n=1^10 (n^2 + 3n - 2)^2$$4\alpha = \sum_{n=1}^{10} 4\left(\frac{n^2 + 3n - 2}{2}\right)^2 = \sum_{n=1}^{10} (n^2 + 3n - 2)^2$$ Expand the squared trinomial: (n^2 + 3n - 2)^2 = n^4 + 9n^2 + 4 + 6n^3 - 4n^2 - 12n = n^4 + 6n^3 + 5n^2 - 12n + 4$$(n^2 + 3n - 2)^2 = n^4 + 9n^2 + 4 + 6n^3 - 4n^2 - 12n = n^4 + 6n^3 + 5n^2 - 12n + 4$$ ### Step 2: Applying given target relation We are given beta = sum_n=1^10 n^4$\beta = \sum_{n=1}^{10} n^4$. So, 4alpha - beta = sum_n=1^10 (n^4 + 6n^3 + 5n^2 - 12n + 4) - sum_n=1^10 n^4$4\alpha - \beta = \sum_{n=1}^{10} (n^4 + 6n^3 + 5n^2 - 12n + 4) - \sum_{n=1}^{10} n^4$ 4alpha - beta = sum_n=1^10 (6n^3 + 5n^2 - 12n + 4)$$4\alpha - \beta = \sum_{n=1}^{10} (6n^3 + 5n^2 - 12n + 4)$$ ### Step 3: Calculating summation limits Evaluate each standard summation up to n=10$n=10$: sum n^3 = (10 times 11 / 2)^2 = 55^2 = 3025$\sum n^3 = (10 \times 11 / 2)^2 = 55^2 = 3025$ sum n^2 = (10 times 11 times 21) / 6 = 385$\sum n^2 = (10 \times 11 \times 21) / 6 = 385$ sum n = (10 times 11) / 2 = 55$\sum n = (10 \times 11) / 2 = 55$ sum 4 = 40$\sum 4 = 40$ 4alpha - beta = 6(3025) + 5(385) - 12(55) + 40$$4\alpha - \beta = 6(3025) + 5(385) - 12(55) + 40$$ = 18150 + 1925 - 660 + 40 = 19455$$= 18150 + 1925 - 660 + 40 = 19455$$ We are given 4alpha - beta = 55k + 40$4\alpha - \beta = 55k + 40$. 19455 = 55k + 40$$19455 = 55k + 40$$ 19415 = 55k$19415 = 55k$ k = frac1941555 = 353$$k = \frac{19415}{55} = 353$$ ### Pattern Recognition Recognizing arithmetic progressions in the first-order differences immediately specifies a quadratic general term An^2+Bn+C$An^2+Bn+C$. Expanding and cancelling highest-order summation terms drastically simplifies standard sums. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

### Related Formula S_n = fracn2[2a + (n-1)d]$$S_n = \frac{n}{2}[2a + (n-1)d]$$ textIf p, q, r text are in G.P. then q^2 = pr$$\text{If } p, q, r \text{ are in G.P. then } q^2 = pr$$ ### Core Logic Let the A.P. be a, a+d, a+2d, dots$a, a+d, a+2d, \dots$ Given a=1$a=1$, the 2^textnd, 8^textth$2^{\text{nd}}, 8^{\text{th}}$, and 44^textth$44^{\text{th}}$ terms are: T_2 = 1 + d$T_2 = 1 + d$ T_8 = 1 + 7d$T_8 = 1 + 7d$ T_44 = 1 + 43d$$T_{44} = 1 + 43d$$ These terms are in G.P., so: (1+7d)^2 = (1+d)(1+43d)$$(1+7d)^2 = (1+d)(1+43d)$$ 1 + 14d + 49d^2 = 1 + 44d + 43d^2$$1 + 14d + 49d^2 = 1 + 44d + 43d^2$$ 6d^2 - 30d = 0 implies 6d(d - 5) = 0$$6d^2 - 30d = 0 \implies 6d(d - 5) = 0$$ Since it is a non-constant A.P., d neq 0$d \neq 0$, so d = 5$d = 5$. Sum of first 20 terms: S_20 = frac202[2(1) + (20-1)5]$$S_{20} = \frac{20}{2}[2(1) + (20-1)5]$$ S_20 = 10[2 + 19(5)] = 10[2 + 95] = 970$$S_{20} = 10[2 + 19(5)] = 10[2 + 95] = 970$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

### Core Logic General term T_r = fracrr^4 - 3r^2 + 1$T_r = \frac{r}{r^4 - 3r^2 + 1}$. Factorize the denominator: r^4 - 3r^2 + 1 = (r^4 - 2r^2 + 1) - r^2 = (r^2 - 1)^2 - r^2$$r^4 - 3r^2 + 1 = (r^4 - 2r^2 + 1) - r^2 = (r^2 - 1)^2 - r^2$$ = (r^2 - r - 1)(r^2 + r - 1)$$= (r^2 - r - 1)(r^2 + r - 1)$$ ### Step 1: Partial Fractions T_r = fracr(r^2 - r - 1)(r^2 + r - 1)$$T_r = \frac{r}{(r^2 - r - 1)(r^2 + r - 1)}$$ Notice that (r^2 + r - 1) - (r^2 - r - 1) = 2r$(r^2 + r - 1) - (r^2 - r - 1) = 2r$. T_r = frac12 left[ frac2r(r^2 - r - 1)(r^2 + r - 1) right] = frac12 left[ frac1r^2 - r - 1 - frac1r^2 + r - 1 right]$$T_r = \frac{1}{2} \left[ \frac{2r}{(r^2 - r - 1)(r^2 + r - 1)} \right] = \frac{1}{2} \left[ \frac{1}{r^2 - r - 1} - \frac{1}{r^2 + r - 1} \right]$$ ### Step 2: Telescoping Sum Sum S = sum_r=1^10 T_r$S = \sum_{r=1}^{10} T_r$. The terms will telescope because the second term for r$r$ is identical to the first term for r+1$r+1$. (Let v_r = r^2 - r - 1$v_r = r^2 - r - 1$, then v_r+1 = (r+1)^2 - (r+1) - 1 = r^2 + 2r + 1 - r - 1 - 1 = r^2 + r - 1$v_{r+1} = (r+1)^2 - (r+1) - 1 = r^2 + 2r + 1 - r - 1 - 1 = r^2 + r - 1$). S = frac12 left[ frac11^2 - 1 - 1 - frac110^2 + 10 - 1 right]$$S = \frac{1}{2} \left[ \frac{1}{1^2 - 1 - 1} - \frac{1}{10^2 + 10 - 1} \right]$$ S = frac12 left[ frac1-1 - frac1109 right] = frac12 left[ -1 - frac1109 right] = -frac55109$$S = \frac{1}{2} \left[ \frac{1}{-1} - \frac{1}{109} \right] = \frac{1}{2} \left[ -1 - \frac{1}{109} \right] = -\frac{55}{109}$$ ### Pattern Recognition Expressions like r^4 + kr^2 + 1$r^4 + kr^2 + 1$ can be factorized by completing the square to create a difference of two squares. This setup invariably leads to a telescoping series. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series