If the coefficients of 2nd, 3rd and 4th terms in the expansion of $(1 + x)^{2 n}$ are in AP, show that $2 n^{2} - 9 n + 7 = 0$

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Solution

It is given that $^{2 n} C_{1},^{2 n} C_{2}$ , and $^{2 n} C_{3}$ are in AP.

$∴ 2 \times^{2 n} C_{2} =^{2 n} C_{1} +^{2 n} C_{3}$

$\Rightarrow 2 = \frac{^{2 n} C_{1}}{^{2 n} C_{2}} + \frac{^{2 n} C_{3}}{^{2 n} C_{2}}$

$\Rightarrow= \frac{4 n}{2 n (2 n - 1)} + \frac{2 n - 3 + 1}{3} ∣ ∣ ∣ ∵ \frac{^{2 n} C_{r}}{^{2 n} C_{r - 1}} = \frac{2 n - r + 1}{r} and r=3 ∣ ∣ ∣$

$\Rightarrow \frac{2}{2 n - 1} + \frac{2 n - 2}{3} = 2$

$\Rightarrow 6 + (2 n - 1) (2 n - 2) = 6 (2 n - 1)$

$\Rightarrow 4 n^{2} - 18 n + 14 = 0 \Rightarrow 2 n^{2} - 9 n + 7 = 0$

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