### Related Formula Sum of binomial coefficients for (x+y)^m$(x+y)^m$ is 2^m$2^m$. Total number of terms is m + 1$m + 1$. Perpendicular distance formula from (x_0, y_0)$(x_0, y_0)$ to ax + by + c = 0$ax + by + c = 0$: d = frac|ax_0 + by_0 + c|sqrta^2 + b^2$d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$ ### Core Logic The power is m = 2n - 3$m = 2n - 3$. Sum of coefficients = 2^2n - 3$= 2^{2n - 3}$. Total terms = 2n - 2$= 2n - 2$. Given Arithmetic Mean: frac2^2n - 32n - 2 = 16 implies frac2^2n - 32(n - 1) = 16 implies 2^2n - 4 = 16(n - 1)$\frac{2^{2n - 3}}{2n - 2} = 16 \implies \frac{2^{2n - 3}}{2(n - 1)} = 16 \implies 2^{2n - 4} = 16(n - 1)$ Testing integer values, n = 5$n = 5$ satisfies the equation perfectly since 2^6 = 64$2^6 = 64$ and 16(5 - 1) = 64$16(5 - 1) = 64$. ### Step 1: Determine Coordinates of Point P Substitute n = 5$n = 5$ into P(2n - 1, n^2 - 4n)$P(2n - 1, n^2 - 4n)$: x_0 = 2(5) - 1 = 9$x_0 = 2(5) - 1 = 9$ y_0 = 5^2 - 4(5) = 5$y_0 = 5^2 - 4(5) = 5$ Thus, P = (9, 5)$P = (9, 5)$.

### Step 2: Calculate Perpendicular Distance Find the distance from (9,5)$(9,5)$ to the line x + y - 8 = 0$x + y - 8 = 0$: d = left|frac9 + 5 - 8sqrt1^2 + 1^2right| = frac6sqrt2 = 3sqrt2$d = \left|\frac{9 + 5 - 8}{\sqrt{1^2 + 1^2}}\right| = \frac{6}{\sqrt{2}} = 3\sqrt{2}$ ### Pattern Recognition Binomial coefficient logic tightly constrains n$n$ to small integers. Use inspection quickly when polynomial-exponential configurations arise. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem Class 11 Mathematics: Straight Lines

### Related Formula General term formula in (a+b)^n$(a+b)^n$: T_r+1 = ^nmathrmC_r a^n-r b^r$T_{r+1} = {}^{n}\mathrm{C}_{r} a^{n-r} b^r$ ### Core Logic The 15^mathrmth$15^{\mathrm{th}}$ term from the beginning corresponds to r=14$r=14$: T_15 = ^nmathrmC_14 left(2^1/3right)^n-14 left(3^-1/3right)^14$T_{15} = {}^{n}\mathrm{C}_{14} \left(2^{1/3}\right)^{n-14} \left(3^{-1/3}\right)^{14}$ The 15^mathrmth$15^{\mathrm{th}}$ term from the end corresponds to the 15^mathrmth$15^{\mathrm{th}}$ term from the beginning if choices are flipped, meaning r = n-14$r = n-14$: T'_15 = ^nmathrmC_n-14 left(2^1/3right)^14 left(3^-1/3right)^n-14$T'_{15} = {}^{n}\mathrm{C}_{n-14} \left(2^{1/3}\right)^{14} \left(3^{-1/3}\right)^{n-14}$ ### Step 1: Find Ratio and Exponents Since ^nmathrmC_14 = ^nmathrmC_n-14${}^{n}\mathrm{C}_{14} = {}^{n}\mathrm{C}_{n-14}$, the combinations cancel in the ratio: fracT_15T'_15 = fracleft(2^1/3right)^n-28left(3^-1/3right)^28-n = left(2^1/3right)^n-28 left(3^1/3right)^n-28 = 6^fracn-283$\frac{T_{15}}{T'_{15}} = \frac{\left(2^{1/3}\right)^{n-28}}{\left(3^{-1/3}\right)^{28-n}} = \left(2^{1/3}\right)^{n-28} \left(3^{1/3}\right)^{n-28} = 6^{\frac{n-28}{3}}$ Given ratio = frac16 = 6^-1$= \frac{1}{6} = 6^{-1}$: 6^fracn-283 = 6^-1 implies fracn-283 = -1 implies n - 28 = -3 implies n = 25$6^{\frac{n-28}{3}} = 6^{-1} \implies \frac{n-28}{3} = -1 \implies n - 28 = -3 \implies n = 25$ ### Step 2: Calculate Binomial Combination ^25mathrmC_3 = frac25 times 24 times 233 times 2 times 1 = 25 times 4 times 23 = 2300${}^{25}\mathrm{C}_{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 25 \times 4 \times 23 = 2300$ ### Pattern Recognition The ratio of symmetric indexed terms from start and end isolates base product factors (a cdot b$a \cdot b$) exclusively as the binomial coefficients perfectly drop out. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

### Related Formula The general term T_r+1$T_{r+1}$ in the expansion of (1+x)^m$(1+x)^m$ is given by: T_r+1 = binommrx^r$T_{r+1} = \binom{m}{r}x^r$ ### Core Logic Identify coefficients A$A$ and B$B$ from the power m = 2n-1$m = 2n-1$ [cite: 3347, 4010]: A = textcoefficient of T_30 = binom2n-129$A = \text{coefficient of } T_{30} = \binom{2n-1}{29}$ B = textcoefficient of T_12 = binom2n-111$B = \text{coefficient of } T_{12} = \binom{2n-1}{11}$ ### Step 1: Set up the ratio equation Use the condition 2A = 5B$2A = 5B$ [cite: 3347, 4010]: 2 cdot binom2n-129 = 5 cdot binom2n-111$2 \cdot \binom{2n-1}{29} = 5 \cdot \binom{2n-1}{11}$ 2 cdot frac(2n-1)!29!(2n-30)! = 5 cdot frac(2n-1)!11!(2n-12)!$2 \cdot \frac{(2n-1)!}{29!(2n-30)!} = 5 \cdot \frac{(2n-1)!}{11!(2n-12)!}$ Cancel (2n-1)!$(2n-1)!$ from both sides: frac229 cdot 28 cdot dots cdot 12 cdot 11! cdot (2n-30)! = frac511! cdot (2n-12)(2n-13) dots (2n-29)(2n-30)!$\frac{2}{29 \cdot 28 \cdot \dots \cdot 12 \cdot 11! \cdot (2n-30)!} = \frac{5}{11! \cdot (2n-12)(2n-13) \dots (2n-29)(2n-30)!}$ frac229 cdot 28 cdot dots cdot 12 = frac5(2n-12)(2n-13) dots (2n-29)$\frac{2}{29 \cdot 28 \cdot \dots \cdot 12} = \frac{5}{(2n-12)(2n-13) \dots (2n-29)}$ ### Step 2: Solve by matching factors Rewriting the relation structurally : By comparing symmetric products or checking valid integers from choices [cite: 4011, 4012]: 2n - 12 = 30 Rightarrow 2n = 42 Rightarrow n = 21$2n - 12 = 30 \Rightarrow 2n = 42 \Rightarrow n = 21$ [cite: 4012, 4013] ### Pattern Recognition Factorial equations can be accelerated by testing options directly into the combinatorial equation 2binom2n-129 = 5binom2n-111$2\binom{2n-1}{29} = 5\binom{2n-1}{11}$ to see which integer matches the ratio demands instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem

### Related Formula The coefficient of the r^textth$r^{\text{th}}$ term in the expansion (1+x)^m$(1+x)^m$ is written as binommr-1$\binom{m}{r-1}$. For three terms in A.P., their values satisfy: 2 cdot T_2 = T_1 + T_3$2 \cdot T_2 = T_1 + T_3$ ### Core Logic Let the total power exponent be m = n + 4$m = n + 4$. The coefficients of the 5^textth$5^{\text{th}}$, 6^textth$6^{\text{th}}$, and 7^textth$7^{\text{th}}$ terms are binomm4$\binom{m}{4}$, binomm5$\binom{m}{5}$, and binomm6$\binom{m}{6}$ respectively. Since they form an arithmetic progression: 2 cdot binomm5 = binomm4 + binomm6$2 \cdot \binom{m}{5} = \binom{m}{4} + \binom{m}{6}$ Rearrange using the recurrence addition identity properties: 4 cdot binomm5 = left[ binomm4 + binomm5 right] + left[ binomm5 + binomm6 right]$4 \cdot \binom{m}{5} = \left[ \binom{m}{4} + \binom{m}{5} \right] + \left[ \binom{m}{5} + \binom{m}{6} \right]$ 4 cdot binomm5 = binomm+15 + binomm+16$4 \cdot \binom{m}{5} = \binom{m+1}{5} + \binom{m+1}{6}$ 4 cdot binomm5 = binomm+26$4 \cdot \binom{m}{5} = \binom{m+2}{6}$ ### Step 1: Solve the Combinatorial Fraction for m$m$ Expand the combinations using factorials: 4 cdot fracm!5!(m-4)! = frac(m+2)!6!(m-4)!$4 \cdot \frac{m!}{5!(m-4)!} = \frac{(m+2)!}{6!(m-4)!}$ Cancel out (m-4)!$(m-4)!$ from both denominators: 4 cdot fracm!120 = frac(m+2)(m+1)m!720$4 \cdot \frac{m!}{120} = \frac{(m+2)(m+1)m!}{720}$ 4 = frac(m+2)(m+1)6$4 = \frac{(m+2)(m+1)}{6}$ 24 = m^2 + 3m + 2 implies m^2 + 3m - 22 = 0$24 = m^2 + 3m + 2 \implies m^2 + 3m - 22 = 0$ Wait, let's re-verify the step using the direct ratio computation: 2 = fracbinomm4binomm5 + fracbinomm6binomm5 = frac5m-4 + fracm-56$2 = \frac{\binom{m}{4}}{\binom{m}{5}} + \frac{\binom{m}{6}}{\binom{m}{5}} = \frac{5}{m-4} + \frac{m-5}{6}$ 2 = frac30 + (m-4)(m-5)6(m-4)$2 = \frac{30 + (m-4)(m-5)}{6(m-4)}$ 12(m-4) = 30 + m^2 - 9m + 20$12(m-4) = 30 + m^2 - 9m + 20$ 12m - 48 = m^2 - 9m + 50 implies m^2 - 21m + 98 = 0$12m - 48 = m^2 - 9m + 50 \implies m^2 - 21m + 98 = 0$ Factor the quadratic equation: (m-7)(m-14) = 0 implies m = 7 text or m = 14$(m-7)(m-14) = 0 \implies m = 7 \text{ or } m = 14$ ### Step 2: Connect back to the problem constraints Since m = n + 4$m = n + 4$: - If m = 14 implies n + 4 = 14 implies n = 10$m = 14 \implies n + 4 = 14 \implies n = 10$ (this is rejected because the problem states n neq 10$n \neq 10$). - If m = 7 implies n + 4 = 7 implies n = 3$m = 7 \implies n + 4 = 7 \implies n = 3$ (this value is accepted). Thus, the binomial expansion exponent is exactly m = 7$m = 7$. ### Step 3: Extract Maximum Binomial Coefficient For an odd exponent power m = 7$m = 7$, the maximum binomial coefficient corresponds to the middle terms: textMax Coefficient = binom73 = binom74 = frac7 cdot 6 cdot 53 cdot 2 cdot 1 = 35$\text{Max Coefficient} = \binom{7}{3} = \binom{7}{4} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35$ ### Pattern Recognition For consecutive binomial coefficients binomnr-1, binomnr, binomnr+1$\binom{n}{r-1}, \binom{n}{r}, \binom{n}{r+1}$ in A.P., the power parameters satisfy the standard identity (n-2r)^2 = n+2$(n-2r)^2 = n+2$, which allows quick calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem