Let the coefficients of three consecutive ter | vedclass

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Question

$( a + b )^{\frac{1}{2}}$
$T_{ r }, T _{ r +1}, T_{ r +2} ightarrow GP$
$	ext { So, } \frac{ T _{ r +1}}{T_{ r }}=\frac{ T _{ r +2}}{T_{ r +1}}$

Vedclass · Accepted Answer

$283$

Vedclass · Answer

$( a + b )^{\frac{1}{2}}$
$T_{ r }, T _{ r +1}, T_{ r +2} ightarrow GP$
$	ext { So, } \frac{ T _{ r +1}}{T_{ r }}=\frac{ T _{ r +2}}{T_{ r +1}}$
$\frac{{ }^{12} C _{ r }}{{ }^{12} C _{ r -1}}=\frac{{ }^{12} C _{ r +1}}{{ }^{12} C _{ r }}$
$\frac{12-r+1}{r}=\frac{12-(r+1)+1}{r+1}$
$(13-r)(r+1)=(12-r)(r)$
$-r+12 r+13=12 r-r^2$
$13=0$
No value of r possible
So $P=0$
$\left(3^{\frac{1}{4}}+4^{\frac{1}{3}}ight)^{12}=\sum{ }^{12} C _{ r }\left(3^{\frac{1}{4}}ight)^{12- r }\left(4^{\frac{1}{3}}ight)^{ r }$
Exponent of $\left(3^{\frac{1}{4}}ight)$ exponent of $\left(4^{\frac{1}{3}}ight)$ term
$\begin{array}{llr}
12 & 0 & 27 \
0 & 12 & 256
\end{array}$
$q=27+256=283$
$p+q=0+283=283$

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