# If the 2nd, 3rd and 4th Terms in the Expansion of (X + A)N Are 240, 720 and 1080 Respectively, Find X, A, N. - Mathematics

If the 2nd, 3rd and 4th terms in the expansion of (x + a)n are 240, 720 and 1080 respectively, find xan.

#### Solution

$\text{ In the expansion of } \left( x + a \right)^n , \text{ the 2nd, 3rd and 4th terms are } ^{n}{}{C}_1 x^{n - 1} a^1 , ^{n}{}{C}_2 x^{n - 2} a^2 \text{ and } ^{n}{}{C}_3 x^{n - 3} a^3 , \ \text{ respectively } .$

$\text{ According to the question } ,$

$^{n}{}{C}_1 x^{n - 1} a^1 = 240$

$^{n}{}{C}_2 x^{n - 2} a^2 = 720$

$^{n}{}{C}_3 x^{n - 3} a^3 = 1080$

$\Rightarrow \frac{^{n}{}{C}_2 x^{n - 2} a^2}{^{n}{}{C}_1 x^{n - 1} a^1} = \frac{720}{240}$

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

$\Rightarrow \frac{a}{x} = \frac{6}{n - 1} . . . \left( 1 \right)$

$\text{ Also } ,$

$\frac{^{n}{}{C}_3 x^{n - 3} a^3}{^{n}{}{C}_2 x^{n - 2} a^2} = \frac{1080}{720}$

$\Rightarrow \frac{n - 2}{3x}a = \frac{3}{2}$

$\Rightarrow \frac{a}{x} = \frac{9}{2n - 4} . . . \left( 2 \right)$

$\text{ Using } \left( 1 \right) \text{ and } \left( 2 \right) \text{ we get }$

$\frac{6}{n - 1} = \frac{9}{2n - 4}$

$\Rightarrow n = 5$

$\text{ Putting in eqn } \left( 1 \right) \text{ we get }$

$\Rightarrow 2a = 3x$

$\text{ Now } , ^{5}{}{C}_1 x^{5 - 1} \left( \frac{3}{2}x \right) = 240$

$\Rightarrow 15 x^5 = 480$

$\Rightarrow x^5 = 32$

$\Rightarrow x = 2$

$\text{ By putting the value of x and n in} \left( 1 \right) \text{ we get}$

$a = 3$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 18 Binomial Theorem
Exercise 18.2 | Q 33 | Page 40