Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
114 views
in Binomial Theorem by (113k points)
closed by
Find the middle terms in the expansion of \(\rm \left(x + \frac 1 x \right)^{10}\)
1. 10C4
2. 10C6
3. 10C5
4. 10C7

1 Answer

0 votes
by (115k points)
selected by
 
Best answer
Correct Answer - Option 3 : 10C5

Concept:

General term: General term in the expansion of (x + y)n is given by

\(\rm {T_{\left( {r\; + \;1} \right)}} = \;{\;^n}{C_r} × {x^{n - r}} × {y^r}\)

 

Middle terms: The middle terms is the expansion of (x + y) n depends upon the value of n.

  • If n is even, then there is only one middle term i.e. \(\rm \left( {\frac{n}{2} + 1} \right){{\rm{\;}}^{th}}\) term is the middle term.
  • If n is odd, then there are two middle terms i.e. \(\rm {\left( {\frac{{n\; + \;1}}{2}} \right)^{th}}\;\)and \(\rm {\left( {\frac{{n\; + \;3}}{2}} \right)^{th}}\) are two middle terms.

 

 

Calculation:

Here, we have to find the middle terms in the expansion of \(\rm \left(x + \frac 1 x \right)^{10}\)

Here n = 10 (n is even number)

∴ Middle term = \(\rm \left( {\frac{n}{2} + 1} \right) = \left( {\frac{10}{2} + 1} \right) =6th\;term\)

T6 = T (5 + 1) = 10C5 × (x) (10 - 5) × \(\rm \left(\frac {1}{x}\right)^5\)

T6 =  10C5

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...