# If the Coefficients of 2nd, 3rd and 4th Terms in the Expansion of (1 + X)2n Are in A.P., Show that 2 N 2 − 9 N + 7 = 0 - Mathematics

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

#### Solution

$\text{ Given }:$                                                                          $(1 + x )^{2n}$
$\text{ Thus, we have: }$
$T_2 = T_{1 + 1}$
$= ^{2n}{}{C}_1 x^1$
$T_3 = T_{2 + 1}$
$= ^{2n}{}{C}_2 x^2$
$T_4 = T_{3 + 1}$
$= ^{2n}{}{C}_3 x^3$
$\text{ We have coefficients of the 2nd, 3rd and 4th terms in AP } .$
$\therefore 2\left( ^{2n}{}{C}_2 \right) = ^t{2n}{}{C}_1 + ^{2n}{}{C}_3$
$\Rightarrow 2 = \frac{^{2n}{}{C}_1}{^{2n}{}{C}_2} + \frac{^{2n}{}{C}_3}{^{2n}{}{C}_2}$
$\Rightarrow 2 = \frac{2}{2n - 1} + \frac{2n - 2}{3}$
$\Rightarrow 12n - 6 = 6 + 4 n^2 - 4n - 2n + 2$
$\Rightarrow 4 n^2 - 18n + 14 = 0$
$\Rightarrow 2 n^2 - 9n + 7 = 0$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 18 Binomial Theorem
Exercise 18.2 | Q 22 | Page 39