Sequences and Series jee mains questions | Sequences and Series Previous Year Question

(4)

Since, $T_{r}, T_{r + 1}$ and $T_{r + 2}$ are in G.P.

So, $\frac{T_{r + 1}}{T_{r}} = \frac{T_{r + 2}}{T_{r + 1}}$

$\Rightarrow \frac{{}^{12}C_{r}}{{}^{12}C_{r - 1}} = \frac{{}^{12}C_{r + 1}}{{}^{12}C_{r}} \Rightarrow \frac{13 - r}{r} = \frac{12 - r}{r + 1}$

$\Rightarrow 12 r - r^{2} = 13 + 13 - r^{2} - r \Rightarrow 13 = 0$ (not possible)

So, p = 0

Now, ${(3^{1 / 4} + 4^{1 / 3})}^{12} = {}^{12}C_{0} {(3^{1 / 4})}^{12} {(4^{1 / 3})}^{0} + {}^{12}C_{0} {(3^{1 / 4})}^{0} {(4^{1 / 3})}^{12}$

$3^{3} + 4^{4} = 27 + 256 = 283 ∴ q = 283$

Thus, p + q = 0 + 283 = 283.

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