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Ex 8.2
Ex 8.2, 2 Important Deleted for CBSE Board 2022 Exams
Ex 8.2, 3 Deleted for CBSE Board 2022 Exams
Ex 8.2, 4 Important Deleted for CBSE Board 2022 Exams
Ex 8.2,5 Deleted for CBSE Board 2022 Exams
Ex 8.2 6 Important Deleted for CBSE Board 2022 Exams
Ex 8.2,7 Deleted for CBSE Board 2022 Exams
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Ex 8.2,9 Deleted for CBSE Board 2022 Exams
Ex 8.2,10 Important Deleted for CBSE Board 2022 Exams
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Ex 8.2,12 Deleted for CBSE Board 2022 Exams
Ex 8.2
Last updated at Jan. 29, 2020 by Teachoo
Ex 8.2, 8 Find the middle terms in the expansions of (π₯/3+9π¦)^10 Number of terms n = 10 Since n is even. There will be one middle term Middle term = π/2 + 1 = 10/2 + 1 = 5 + 1 = 6th term. Hence we need to find 6th term. i.e. T6 We know that general term of (a + b)n is Tr + 1 = nCr (a)n-r . (b)r Finding 6th term i.e. T6 = T5 + 1 , Hence r = 5 Putting n = 5 , a = π₯/3 & b = 9y and r = 1 in (1) T5 + 1 = 10C5 (π₯/3)^(10 β 5) . (9y)5 T6 = 10C5 (π₯/3)^5 . (9)5 . y5 = 10!/5!(10 β 5)! . (π₯/3)^5. (32) 5 . y5 = 10!/5!5! . π₯5/35 . 310 . y5 = (10 Γ 9 Γ 8 Γ 7 Γ 6 Γ 5!)/(5 Γ 4 Γ 3 Γ 2 Γ 1 Γ5!) . x5 .310/35 . y5 = (10 Γ 9 Γ 8 Γ 7 Γ 6)/(5 Γ 4 Γ 3 Γ 2 Γ 1 ) . x5 . 35 . y5 = (10 Γ 9 Γ 8 Γ 7 Γ 6 Γ35)/(5 Γ 4 Γ 3 Γ 2 Γ 1 ) . x5 . y5 = 61236 x5 y5 Hence, middle term is 61236 x5 y5