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If Rth Term is the Middle Term in the Expansion of ( X 2 − 1 2 X ) 20 Then ( R + 3 ) T H Term Is(A) 20 C 14 ( X 2 14 ) (B) 20 C 12 X 2 2 − 12 (C) − T 20 C 7 X , 2 − 13 (D) None of These

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Question

If rth term is the middle term in the expansion of \[\left( x^2 - \frac{1}{2x} \right)^{20}\]  then \[\left( r + 3 \right)^{th}\]  term is

 

 

Options

  •  \[^{20}{}{C}_{14} \left( \frac{x}{2^{14}} \right)\]

     

  •   \[^{20}{}{C}_{12} x^2 2^{- 12}\]

     

  • \[- ^t{20}{}{C}_7 x, 2^{- 13}\]

     

  •  none of these

     
MCQ
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Solution

 \[- ^t{20}{}{C}_7 x, 2^{- 13}\]
Here n is even
So, The middle term in the given expansion is
\[\left( \frac{20}{2} + 1 \right)\text{ th = 11th term} \] 
Therefore, (r + 3)th term is the 14th term.
\[T_{14} = ^{20}{}{C}_{13} ( x^2 )^{20 - 13} \left( \frac{- 1}{2x} \right)^{13} \]
`= \left( - 1 \right)^{13} "^20C_{13} \frac{x^{14 - 13}}{2^{13}}`
\[ = - ^{20}{}{C}_7 x 2^{- 13}\]

 

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Chapter 18: Binomial Theorem - Exercise 18.4 [Page 48]

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R.D. Sharma Mathematics [English] Class 11
Chapter 18 Binomial Theorem
Exercise 18.4 | Q 30 | Page 48

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