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Misc 8 - Find n, if ratio of fifth term from beginning to end

Misc  8 - Chapter 8 Class 11 Binomial Theorem - Part 2
Misc  8 - Chapter 8 Class 11 Binomial Theorem - Part 3 Misc  8 - Chapter 8 Class 11 Binomial Theorem - Part 4 Misc  8 - Chapter 8 Class 11 Binomial Theorem - Part 5 Misc  8 - Chapter 8 Class 11 Binomial Theorem - Part 6 Misc  8 - Chapter 8 Class 11 Binomial Theorem - Part 7


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Misc 8 Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (∜2+ 1/∜3)^𝑛 is √6 : 1 We know that General term of expansion (a + b)n Tr + 1 = nCr an – rbr Fifth term from beginning We need to calculate T5 = T4 + 1 Putting r = 4 , a = ∜2 , b = (1/∜3) T4 + 1 = nC4(∜2)n – 4.(1/∜3)4 T5 = nC4 γ€–((2)γ€—^(1/4))n – 4 .(1/3^(1/4) )4 T5 = nC4 γ€–(2)γ€—^(1/4 Γ— (𝑛 βˆ’ 4)) .(1/3^(1/4) )4 = nCr γ€–(2)γ€—^(1/4 Γ—(π‘›βˆ’4)) .(1/3) Finding 5th term from end We know that (a + b)n = nCo an bo +nC1 an–1 b1 +……..+ nCn–1 (a)n – (n–1) .bn–1 +nCn a0 bn = an + nC1 an–1b1 + ………………………+ nC1 a1bn–1 + bn = bn + nC1 a1 bn–1 +…………………+ nC1 an–1 b1 + an Hence, 5th term from end = (n – 3)th term from beginning Another method, rth term from end = (n – r + 2)th term from stating 5th term from end = (n – 5 + 2)th term from stating = (n – 3)th term from beginning Now we need to calculate (n – 3)th term of expansion (√(πŸ’&𝟐) + 𝟏/√(πŸ’&πŸ‘) )r Tr + 1 = nCr an – rbr Putting r = (n – 3) – 1 = n – 4 a = ∜2 , b = 1/∜3 T(n – 4 + 1) = nCn – 4 (∜2)^(𝑛 βˆ’(π‘›βˆ’4)) . (1/∜3)^(𝑛 βˆ’ 4) = nCn – 4 [γ€–(2)γ€—^(1/4) ]^"n – n + 4" . (1/3^(1/4) )^(𝑛 βˆ’ 4) = nCn – 4 [γ€–(2)γ€—^(1/4) ]^"4" . (3^((βˆ’1)/4) )^(𝑛 βˆ’ 4) = nCn – 4 γ€–(2)γ€—^1 .(1/3)^((𝑛 βˆ’ 4)/4) = nCn – 4 γ€–(2)γ€—^(4/4) .3^((βˆ’(𝑛 βˆ’ 4))/4) = nCn – 4 γ€–(2)γ€—^1 .(1/3)^((𝑛 βˆ’ 4)/4) Given that Ratio fifth term from beginning fifth term from end is √6 : 1 (πΉπ‘–π‘“π‘‘β„Ž π‘‘π‘’π‘Ÿπ‘š π‘“π‘Ÿπ‘œπ‘š 𝑏𝑒𝑔𝑖𝑛𝑛𝑖𝑛𝑔 )/(πΉπ‘–π‘“π‘‘β„Ž π‘‘π‘’π‘Ÿπ‘š π‘“π‘Ÿπ‘œπ‘š 𝑒𝑛𝑑 ) = √6/1 (〖𝑛𝐢〗_4 2^((π‘›βˆ’4)/4) . 1/3)/(〖𝑛𝐢〗_(π‘›βˆ’4) 2 . (1/3)^((π‘›βˆ’4)/4) ) = √6/1 (〖𝑛𝐢〗_4 )/(〖𝑛𝐢〗_(π‘›βˆ’4) ) . (2)^((𝑛 βˆ’ 4)/4 βˆ’1 ) (1/3)^( 1 βˆ’ ((𝑛 βˆ’ 4)/4) ) = √6/1 (〖𝑛𝐢〗_4 )/(〖𝑛𝐢〗_(π‘›βˆ’4) ) . 2^((𝑛 βˆ’ 4 βˆ’ 4)/4 ). (1/3)^((4 βˆ’ 𝑛 + 4)/4) = √6/1 (〖𝑛𝐢〗_4 )/(〖𝑛𝐢〗_(π‘›βˆ’4) ). 2^((𝑛 βˆ’ 8)/4 ). (1/3)^((8 βˆ’ 𝑛)/4) = √6/1 (〖𝑛𝐢〗_4 )/(〖𝑛𝐢〗_(π‘›βˆ’4) ). 2^((𝑛 βˆ’ 8)/4 ). (1/3)^((βˆ’(𝑛 βˆ’ 8))/4) = √6/1 (〖𝑛𝐢〗_4 )/(〖𝑛𝐢〗_(π‘›βˆ’4) ). 2^((π‘›βˆ’8)/4). (3)^((𝑛 βˆ’ 8)/4) = √6/1 (〖𝑛𝐢〗_4 )/(〖𝑛𝐢〗_(π‘›βˆ’4) ). γ€–(2 Γ— 3)γ€—^((𝑛 βˆ’ 8)/4). = 6^(1/2) (〖𝑛𝐢〗_4 )/(〖𝑛𝐢〗_(π‘›βˆ’4) ). 6^((𝑛 βˆ’ 8)/4)= 6^(1/2) Comparing powers of 6 (𝑛 βˆ’ 8)/4 = 1/2 n – 8 = 4/2 n – 8 = 2 n = 2 + 8 n = 10 Hence, n = 10

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.