The coefficients of $x^{16}$ and $x^{17}$ in $(a + b x)^{101}$ are equal. If a and b are natural numbers, then the least possible value of $a$ is

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Solution

The correct option is B $5$
It is given that coefficients of $T_{17}$ and $T_{18}$ are equal
Hence
$^{101} C_{16} a^{85} b^{16} =^{101} C_{17} a^{84} b^{17}$
$\frac{101! a}{16! 85!} = \frac{101! b}{17! 84!}$
$\frac{a}{85} = \frac{b}{17}$
$a = 5 b$
Hence least value of a is 5 given that $a, b ϵ N$

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