Check sibling questions

Ex 8.2, 7 - Find middle term of (3 - x3/6)7 - Chapter 8 - Ex 8.2

Ex 8.2,7 - Chapter 8 Class 11 Binomial Theorem - Part 2
Ex 8.2,7 - Chapter 8 Class 11 Binomial Theorem - Part 3
Ex 8.2,7 - Chapter 8 Class 11 Binomial Theorem - Part 4

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Ex 8.2, 7 Find the middle terms in the expansions of ("3 – " π‘₯3/6)^7 Number of terms = n = 7 Since n is odd there will be two middle termx ((𝑛 + 1)/2)^π‘‘β„Ž term & ((𝑛 + 1)/2+1)^π‘‘β„Ž term Hence we need to find 4th and 5th term i.e. T4 and T5 We know that general term of (a + b)n is Tr + 1 = nCr an – r br = ((7 + 1)/2)^π‘‘β„Ž term = 4th term = ((7 + 1)/2)^π‘‘β„Ž term = 4th term For ("3 – " π’™πŸ‘/πŸ”)^πŸ• Putting a = 3 , b = ((βˆ’π‘₯3)/6) , n = 7 Tr + 1 = 7Cr (3)7 – r ("–" π‘₯3/6)^π‘Ÿ Tr + 1 = 7Cr (3)7 – r (–1)r (π‘₯3/6)^π‘Ÿ Tr + 1 = 7Cr (3)7 – r (–1)r (π‘₯3/6)^π‘Ÿ Tr + 1 = 7Cr (3)7 – r (–1)r (1/6)^π‘Ÿ π‘₯^3π‘Ÿ Finding 4th term Tr + 1 = 7Cr (3)7 – r (–1)r (1/6)^π‘Ÿ π‘₯^3π‘Ÿ Finding 4th term i.e. T4 = T3 + 1 , Hence r = 3 T3+1 = 7C3 (3)7 – 3 (–1)3 (1/6)^3 π‘₯^(3(3)) T4 = 7C3 . 34 . (βˆ’1) . 1/63 . (x)9 = – 7!/3!(7 βˆ’ 3)! . 34. 1/((2 Γ— 3)3) . π‘₯9 = – 7!/(3! 4!) . 34 . 1/(23 . 33) . π‘₯9 = – ((7 Γ— 6 Γ— 5 Γ— 4!)/(3 Γ— 2 Γ— 1 Γ— 4!)) . 34/33 . π‘₯9/23 = (βˆ’πŸπŸŽπŸ“)/πŸ– x9 Finding 5th term Tr + 1 = 7Cr (3)7 – r (–1)r (1/6)^π‘Ÿ π‘₯^3π‘Ÿ Finding 5th term i.e. T5 = T4 + 1 , Hence r = 4 T4+1 = 7C4 (3)7 - 4 (–1)4 (1/6)^4 π‘₯^(3(4)) = 7C4 . 33 . 1/64 . (x)12 = 7!/4!(7 βˆ’ 4)! . 33 . 1/(2 Γ— 3)4 . x12 = 7!/(3! 4!) . 33 . 1/(24 . 34) . π‘₯12 = ((7 Γ— 6 Γ— 5 Γ— 4!)/(3 Γ— 2 Γ— 1 Γ— 4!)) . 34/34 . π‘₯12/24 = πŸ‘πŸ“/πŸ’πŸ– x12 Hence, the middle terms are (βˆ’πŸπŸŽπŸ“)/πŸ– x9 & πŸ‘πŸ“/πŸ’πŸ– x12

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.