Advertisement Remove all ads

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.

Advertisement Remove all ads

Solution

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

\[\text{ According to the question } , \]

\[ ^{n}{}{C}_1 x^{n - 1} a^1 = 240 \]

\[ ^{n}{}{C}_2 x^{n - 2} a^2 = 720\]

\[^{n}{}{C}_3 x^{n - 3} a^3 = 1080\]

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

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

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

\[\text{ Also } , \]

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

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

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

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

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

\[ \Rightarrow n = 5\]

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

\[ \Rightarrow 2a = 3x\]

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

\[ \Rightarrow 15 x^5 = 480\]

\[ \Rightarrow x^5 = 32\]

\[ \Rightarrow x = 2\]

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

\[a = 3\]

 

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

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

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×