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If the $17^{\text {th }}$ and the $18^{\text {th }}$ terms in the expansion of $(2+a)^{50}$ are equal, then the coefficient of $x^{35}$ in the expansion of $(a+x)^2$ is
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Verified Answer
The correct answer is:
-36
Given,
$17^{\text {th }}$ and $18^{\text {th }}$ terms in the expansion $(2+a)^{50}$ are equal
$$
\begin{array}{rlrl}
& \therefore & \mathrm{T}_{17} & =\mathrm{T}_{18} \\
\Rightarrow & { }^{50} \mathrm{C}_{16}(2)^{34}(a)^{16} & ={ }^{50} \mathrm{C}_{17}(2)^{33}(a)^{17} \\
\Rightarrow & a =\frac{{ }^{50} C_{16}}{{ }^{50} C_{17}} \times 2=1
\end{array}
$$
$\therefore$ Now, coefficient of $x^{35}$ in the expansion
$$
(1+x)^{-2}=-36
$$
$17^{\text {th }}$ and $18^{\text {th }}$ terms in the expansion $(2+a)^{50}$ are equal
$$
\begin{array}{rlrl}
& \therefore & \mathrm{T}_{17} & =\mathrm{T}_{18} \\
\Rightarrow & { }^{50} \mathrm{C}_{16}(2)^{34}(a)^{16} & ={ }^{50} \mathrm{C}_{17}(2)^{33}(a)^{17} \\
\Rightarrow & a =\frac{{ }^{50} C_{16}}{{ }^{50} C_{17}} \times 2=1
\end{array}
$$
$\therefore$ Now, coefficient of $x^{35}$ in the expansion
$$
(1+x)^{-2}=-36
$$
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