Numerically the greatest term in the expansion of $(2 + 3 x)^{12}$ , when $x = \frac{5}{6}$ is

^{12} C_{7} 2^{4} {(\frac{5}{2})}^{7}

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^{12} C_{5} 2^{5} {(\frac{5}{2})}^{7}

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^{12} C_{5} 2^{5} {(\frac{5}{4})}^{7}

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^{12} C_{5} 2^{3} {(\frac{5}{2})}^{5}

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Solution

The correct option is C $^{12} C_{5} 2^{5} {(\frac{5}{2})}^{7}$
Let $r^{t h}$ and ${(r + 1)}^{t h}$ teem denoted by $T r, T r + 1$ respectively
we know that in the expansion ${(x + a)}^{n}$ we have $\frac{T r + 1}{T r} = \frac{n - r + 1}{r} \frac{a}{x}$
$∴$ in ${(2 + 3 x)}^{12},$ we have
$\frac{T r + 1}{T r} = \frac{12 - r + 1}{r} (\frac{3 x}{2}) = \frac{12 - r + 1}{r} (\frac{3}{2} \times \frac{5}{6}) = \frac{65 - 5 r}{4 r}$
now $T r + 1 > T r \frac{65 - 5 r}{4 r} \Rightarrow r < \frac{65}{9} = 7 \frac{2}{9}$
Again $T r + 1 = T r \Rightarrow r = 7 \frac{2}{9}$ and $T r + 1 < T r \Rightarrow r > 7 \frac{2}{9}$
so up to $7$ for all values of $r, T r + 1 > T r .$ and for all values of $r \geq 8 T r + 1 < T r$
Hence the greatest term in $T r + 1$ where $T r = 7$
$∴ T_{8}$ is greatest term now, $T_{8} = T_{7 + 1} =^{12} C_{7} 2^{12 - r} {(3 x)}^{7}$
$=^{12} C_{7} 2^{5} {(\frac{3 \times 5}{6})}^{7} (∵ x = \frac{5}{6}) =^{12} C_{7} 2^{5} {(\frac{5}{2})}^{7}$

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