# Answer the following: Find the middle term (s) in the expansion of (3x22-13x)9 - Mathematics and Statistics

Sum

Find the middle term (s) in the expansion of ((3x^2)/2 - 1/(3x))^9

#### Solution

There are 10 terms in the expansion of ((3x^2)/2 - 1/(3x))^9.

∴ middle terms are t5 and t6.

We know that, in the expansion of (a+ b)n,

tr+1 = nCr an-r br

Here a = (3x^2)/2, b = -1/(3x), n = 9

For t5, r = 4 and for t6, r = 5

∴ t5 = ""^9"C"_4 ((3x^2)/2)^(9 - 4) (-1/(3x))^4

= (9 xx 8 xx 7 xx 6)/(1 xx 2 xx 3 xx 4) xx (3^5.x^10)/32 xx 1/(3^4.x^4)

= (189x^6)/16

and t6 = ""^9"C"_5 ((3x^2)/2)^(9 - 5) (-1/(3x))^5

= (9 xx 8 xx 7 xx 6 xx 5)/(1 xx 2 xx 3 xx 4 xx 5) xx (3^4.x^8)/16 xx (-1)/(3^5.x^5)

= (-21x^3)/8

∴ the middle terms are (189x^6)/16 and (-21x^3)/8.

Concept: Middle term(s) in the expansion of (a + b)n
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Miscellaneous Exercise 4 | Q II. (8) (iv) | Page 85