### Related Formula General term of binomial expansion (x + y)^n$(x + y)^n$: T_k+1 = binomnk x^n-k y^k$T_{k+1} = \binom{n}{k} x^{n-k} y^k$ Condition for three terms A, B, C$A, B, C$ to be in G.P.: B^2 = A cdot C$B^2 = A \cdot C$ ### Core Logic Part 1: Coefficients of T_r, T_r+1, T_r+2$T_r, T_{r+1}, T_{r+2}$ are binom12r-1, binom12r, binom12r+1$\binom{12}{r-1}, \binom{12}{r}, \binom{12}{r+1}$. Since they are in G.P.: left[binom12rright]^2 = binom12r-1 cdot binom12r+1$\left[\binom{12}{r}\right]^2 = \binom{12}{r-1} \cdot \binom{12}{r+1}$ fracbinom12rbinom12r-1 = fracbinom12r+1binom12r implies frac12-r+1r = frac12-rr+1$\frac{\binom{12}{r}}{\binom{12}{r-1}} = \frac{\binom{12}{r+1}}{\binom{12}{r}} \implies \frac{12-r+1}{r} = \frac{12-r}{r+1}$ (13-r)(r+1) = r(12-r) implies 13r + 13 - r^2 - r = 12r - r^2$(13-r)(r+1) = r(12-r) \implies 13r + 13 - r^2 - r = 12r - r^2$ 12r + 13 = 12r implies 13 = 0 quad (textNot possible)$12r + 13 = 12r \implies 13 = 0 \quad (\text{Not possible})$ Thus, no real integer solution for r$r$ exists, so p = 0$p = 0$. ### Step 1: Calculate Rational Terms Sum q Part 2: Rational terms in expansion of (3^1/4 + 4^1/3)^12$(3^{1/4} + 4^{1/3})^{12}$. General term: T_k+1 = binom12k (3^1/4)^12-k (4^1/3)^k = binom12k 3^frac12-k4 4^frack3$T_{k+1} = \binom{12}{k} (3^{1/4})^{12-k} (4^{1/3})^k = \binom{12}{k} 3^{\frac{12-k}{4}} 4^{\frac{k}{3}}$ For the term to be rational, frac12-k4$\frac{12-k}{4}$ and \frac{k}{3} must both be integers: - k$k$ must be a multiple of 3: k in \0, 3, 6, 9, 12\$k \in \{0, 3, 6, 9, 12\}$ - 12-k$12-k$ must be a multiple of 4, so k$k$ must be a multiple of 4: k in \0, 4, 8, 12\$k \in \{0, 4, 8, 12\}$ The common values for k$k$ are k = 0$k = 0$ and k = 12$k = 12$. - At k = 0$k = 0$: T_1 = binom120 3^3 4^0 = 1 times 27 times 1 = 27$T_1 = \binom{12}{0} 3^3 4^0 = 1 \times 27 \times 1 = 27$ - At k = 12$k = 12$: T_13 = binom1212 3^0 4^4 = 1 times 1 times 256 = 256$T_{13} = \binom{12}{12} 3^0 4^4 = 1 \times 1 \times 256 = 256$ Sum of rational terms q = 27 + 256 = 283$q = 27 + 256 = 283$. ### Step 2: Final Combination p + q = 0 + 283 = 283$p + q = 0 + 283 = 283$ ### Pattern Recognition To find common values for divisibility constraints, look for multiples of the least common multiple textlcm(3, 4) = 12$\text{\lcm}(3, 4) = 12$ within the range [0, 12]$[0, 12]$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem