# Find the middle term (terms) in the expansion of (3x-x36)9 - Mathematics

Sum

Find the middle term (terms) in the expansion of (3x - x^3/6)^9

#### Solution

Given expression is (3x - x^3/6)^9

Number of terms = 9 + 1 = 10  ....(even)

∴ Middle terms are n^"th"/2 term and (n/2 + 1)^"th" term

= 10^"th"/2

= 5th term and (5 + 1) = 6th term

General Term "T"_(r + 1) = ""^n"C"_r  x^(n - r) y^r

∴ T5 = "T"_(4 + 1)

= ""^9"C"_4  (3x)^(9 - 4)  (- x^3/6)^4

= ""^9"C"_4  (3)^5 * x^5  (-1/6)^4 * x^12

= (9 xx 8 xx 7 xx 6)/(4 xx 3 xx 2 xx 1) xx (3 xx 3 xx 3 xx 3 xx 3)/(6 xx 6 xx 6 xx 6) x^17

= 189/8 x^17

Now, T6 = T5+1

= ""^9"C"_5  (3x)^(9 - 5) (- x^3/6)^5

= ""^9"C"_5  (3)^4 x^4 (- 1/6)^5 * x^15

= (9 xx 8 xx 7 xx 6 xx 5)/(5 xx 4 xx 3 xx 2 xx 1) (3)^4 (- 1/6)^5 * x^19

=  - 21/16 x^19

Hence, the required middle terms are 189/8 x^17 and - 21/16 x^19

Concept: General and Middle Terms
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Chapter 8: Binomial Theorem - Exercise [Page 142]

#### APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 8 Binomial Theorem
Exercise | Q 5.(ii) | Page 142

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