If the sum of the first 10 terms of the series frac4 cdot 11 + 4 cdot 1^4 + frac4 cdot 21 + 4 cdot 2^4 + frac4 cdot 31 + 4 cdot 3^4 + dots$\frac{4 \cdot 1}{1 + 4 \cdot 1^4} + \frac{4 \cdot 2}{1 + 4 \cdot 2^4} + \frac{4 \cdot 3}{1 + 4 \cdot 3^4} + \dots$ is fracmathrmmmathrmn$\frac{\mathrm{m}}{\mathrm{n}}$, where gcd(mathrmm, mathrmn) = 1$\gcd(\mathrm{m}, \mathrm{n}) = 1$, then mathrmm + mathrmn$\mathrm{m} + \mathrm{n}$ is equal to ____________.

### 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

### Related Formula - General term of an AP: T_n = a + (n-1)d$T_n = a + (n-1)d$ - Sum of n$n$ terms of an AP: S_n = fracn2[2a + (n-1)d]$S_n = \frac{n}{2}[2a + (n-1)d]$ - The common difference of a series of common terms between two APs is given by the least common multiple of their respective common differences: d_textcommon = textLCM(d_1, d_2)$$d_{\text{common}} = \text{LCM}(d_1, d_2)$$ ### Core Logic Let's analyze both arithmetic progressions: - AP 1: 3, 7, 11, 15, dots, 403 implies$3, 7, 11, 15, \dots, 403 \implies$ First term a_1 = 3$a_1 = 3$, common difference d_1 = 4$d_1 = 4$. - AP 2: 2, 5, 8, 11, dots, 404 implies$2, 5, 8, 11, \dots, 404 \implies$ First term a_2 = 2$a_2 = 2$, common difference d_2 = 3$d_2 = 3$. By observation, the first identical value appearing in both series is 11$11$. Therefore, the new common AP has: - First term a = 11$a = 11$ - Common difference d = textLCM(4, 3) = 12$d = \text{LCM}(4, 3) = 12$ ### Step 1: Determine the Number of Common Terms The last term T_n$T_n$ of the common AP cannot exceed the boundary upper limits of either individual series (i.e., le 403$\le 403$): T_n = 11 + (n-1)12 le 403$$T_n = 11 + (n-1)12 \le 403$$ 12(n-1) le 392$12(n-1) \le 392$ n-1 le 32.66 implies n = 33$$n-1 \le 32.66 \implies n = 33$$ ### Step 2: Calculate the Series Sum Using the AP summation formula for 33$33$ terms: S_33 = frac332 left[ 2(11) + (33-1)12 right]$$S_{33} = \frac{33}{2} \left[ 2(11) + (33-1)12 \right]$$ S_33 = frac332 left[ 22 + 32 times 12 right]$$S_{33} = \frac{33}{2} \left[ 22 + 32 \times 12 \right]$$ S_33 = frac332 left[ 22 + 384 right]$$S_{33} = \frac{33}{2} \left[ 22 + 384 \right]$$ S_33 = frac332 times 406 = 33 times 203 = 6699$$S_{33} = \frac{33}{2} \times 406 = 33 \times 203 = 6699$$ ### Pattern Recognition Sees: Overlapping arithmetic progression series elements. Shortcut: Once the first matching number and the LCM are calculated, the maximum term inequality a + (n-1)d le min(L_1, L_2)$a + (n-1)d \le \min(L_1, L_2)$ directly maps out the total number of terms cleanly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula If x, y, z$x, y, z$ are in A.P., then 2y = x + z$2y = x + z$. ### Core Logic From the first sequence condition: 2log_e b = log_e a + log_e c implies log_e b^2 = log_e(ac) implies b^2 = ac quad dots (i)$$2\log_e b = \log_e a + \log_e c \implies \log_e b^2 = \log_e(ac) \implies b^2 = ac \quad \dots (i)$$ From the second sequence condition, the components are log_eleft(fraca2bright), log_eleft(frac2b3cright), log_eleft(frac3caright)$\log_e\left(\frac{a}{2b}\right), \log_e\left(\frac{2b}{3c}\right), \log_e\left(\frac{3c}{a}\right)$: 2log_eleft(frac2b3cright) = log_eleft(fraca2bright) + log_eleft(frac3caright)$$2\log_e\left(\frac{2b}{3c}\right) = \log_e\left(\frac{a}{2b}\right) + \log_e\left(\frac{3c}{a}\right)$$ left(frac2b3cright)^2 = fraca2b times frac3ca = frac3c2b$$\left(\frac{2b}{3c}\right)^2 = \frac{a}{2b} \times \frac{3c}{a} = \frac{3c}{2b}$$ frac4b^29c^2 = frac3c2b implies 8b^3 = 27c^3 implies fracbc = frac32 quad dots (ii)$$\frac{4b^2}{9c^2} = \frac{3c}{2b} \implies 8b^3 = 27c^3 \implies \frac{b}{c} = \frac{3}{2} \quad \dots (ii)$$ ### Step 1: Finding Ratios Substituting c = frac2b3$c = \frac{2b}{3}$ into equation (i): b^2 = a left(frac2b3right) implies b = frac2a3 implies fracab = frac32$$b^2 = a \left(\frac{2b}{3}\right) \implies b = \frac{2a}{3} \implies \frac{a}{b} = \frac{3}{2}$$ Thus, consolidating all parts: a : b = 9 : 6$a : b = 9 : 6$ b : c = 6 : 4$b : c = 6 : 4$ a : b : c = 9 : 6 : 4$$a : b : c = 9 : 6 : 4$$ ### Pattern Recognition Logarithmic A.P. strings immediately translate to simple geometric proportions inside the core arguments via logarithmic properties (2log x = log x^2$2\log x = \log x^2$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series

### Related Formula 2a_n = a_n+1 + a_n+2$$2a_n = a_{n+1} + a_{n+2}$$ ### Core Logic Let the terms of the Geometric Progression have a common ratio r$r$. Substituting the geometric forms into the arithmetic mean relationship: 2(a r^n-1) = a r^n + a r^n+1$$2(a r^{n-1}) = a r^n + a r^{n+1}$$ Dividing out non-zero fields a r^n-1$a r^{n-1}$: 2 = r + r^2 implies r^2 + r - 2 = 0$$2 = r + r^2 \implies r^2 + r - 2 = 0$$ (r + 2)(r - 1) = 0$$(r + 2)(r - 1) = 0$$ Since a_2 neq a_1$a_2 \neq a_1$, we have r neq 1$r \neq 1$. Thus, the common ratio is r = -2$r = -2$. ### Step 1: Evaluating the Target Partial Difference We need to evaluate: S_20 - S_18 = T_19 + T_20$$S_{20} - S_{18} = T_{19} + T_{20}$$ T_19 + T_20 = a r^18 + a r^19 = a r^18(1 + r)$$T_{19} + T_{20} = a r^{18} + a r^{19} = a r^{18}(1 + r)$$ Substituting a = frac18$a = \frac{1}{8}$ and r = -2$r = -2$: T_19 + T_20 = frac18 (-2)^18 (1 - 2) = frac12^3 cdot 2^18 cdot (-1) = -2^15$$T_{19} + T_{20} = \frac{1}{8} (-2)^{18} (1 - 2) = \frac{1}{2^3} \cdot 2^{18} \cdot (-1) = -2^{15}$$ ### Pattern Recognition Partial sum differences simplify into standard standalone term values (S_n - S_n-2 = T_n + T_n-1$S_n - S_{n-2} = T_n + T_{n-1}$). This eliminates the need to apply long fraction sum formats. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series