If the coefficient of three successive terms in expansion of $(1 + x)^{n}$ is 45, 120 and 210. Find n.

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Solution

Let the terms be $^{n} C_{r - 1},^{n} C_{r},^{n} C_{r + 1}$
$\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{120}{45}$
$\frac{\frac{n!}{(n - r)! r!}}{\frac{n!}{(n - r + 1)! (r - 1)!}} = \frac{8}{3}$

$\frac{n - r + 1}{r} = \frac{8}{3}$
$3 n - 3 r + 3 - 8 r = 0 \Rightarrow 3 n - 11 r = - 3............. (1)$

$\frac{^{n} C_{r + 1}}{^{n} C_{r}} = \frac{210}{120}$

$\frac{\frac{n!}{(n - r - 1)! (r + 1)!}}{\frac{n!}{(n - r)! r!}} = \frac{7}{4}$

$\frac{n - r}{r + 1} = \frac{7}{4}$
$4 n - 11 r = 7.............. (2)$
Subtracting (2) from (1)
$3 n - 11 r - (4 n - 11 r) = - 10$
$n = 10$

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