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QJEEM25Ma89d7: JEE Main 2025 Mathematics - Class 11 JEEM25Ma89d7 Solution: Let the coefficients of three consecutive terms , and in the…

Let the coefficients of three consecutive terms TrT_r , Tr+1T_{r+1} and Tr+2T_{r+2} in the binomial expansion of (a+b)12(a + b)^{12} be in a G.P. and let pp be the number of all possible values of rr . Let qq be the sum of all rational terms in the binomial expansion of (34+43)12(\sqrt[4]{3}+\sqrt[3]{4})^{12} . Then p+qp + q is equal to:

Multiple Choice Options

  • A 295
  • B 283
  • C 299
  • D 287

Step-by-Step Verified Solution

amp;(a+b)12amp; Tr, Tr+1, Tr+2→GPamp; So, Tr+1 Tr=Tr+2 Tr+1amp;12Cr12Cr−1=12Cr+112Cramp;12−r+1r=12−(r+1)+1r+1amp;(13−r)(r+1)=(12−r)(r)amp;−r+12r+13=12r−r2amp;13=0\begin{aligned}& (\mathrm{a}+\mathrm{b})^{\frac{1}{2}} \\& \mathrm{~T}_{\mathrm{r}}, \mathrm{~T}_{\mathrm{r}+1}, \mathrm{~T}_{\mathrm{r}+2} \rightarrow \mathrm{GP} \\& \text { So, } \frac{\mathrm{T}_{\mathrm{r}+1}}{\mathrm{~T}_{\mathrm{r}}}=\frac{\mathrm{T}_{\mathrm{r}+2}}{\mathrm{~T}_{\mathrm{r}+1}} \\& \frac{{ }^{12} \mathrm{C}_{\mathrm{r}}}{{ }^{12} \mathrm{C}_{\mathrm{r}-1}}=\frac{{ }^{12} \mathrm{C}_{\mathrm{r}+1}}{{ }^{12} \mathrm{C}_{\mathrm{r}}} \\& \frac{12-\mathrm{r}+1}{\mathrm{r}}=\frac{12-(\mathrm{r}+1)+1}{\mathrm{r}+1} \\& (13-\mathrm{r})(\mathrm{r}+1)=(12-\mathrm{r})(\mathrm{r}) \\& -\mathrm{r}+12 \mathrm{r}+13=12 \mathrm{r}-\mathrm{r}^2 \\& 13=0\end{aligned}

No value of r possible

So P=0\mathrm{P}=0

(314+413)12=∑12Cr(314)12−r(413)r\left(3^{\frac{1}{4}}+4^{\frac{1}{3}}\right)^{12}=\sum{ }^{12} C_r\left(3^{\frac{1}{4}}\right)^{12-\mathrm{r}}\left(4^{\frac{1}{3}}\right)^{\mathrm{r}}

Exponent of (314)\left(3^{\frac{1}{4}}\right) exponent of (413)\left(4^{\frac{1}{3}}\right) term

12amp;0amp;270amp;12amp;256q=27+256=283amp;amp;p+q=0+283=283amp;amp;\begin{array}{ccc}12 & 0 & 27 \\0 & 12 & 256 \\\mathrm{q}=27+256=283 & & \\\mathrm{p}+\mathrm{q}=0+283=283 & &\end{array}

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