# Misc 8

Last updated at March 8, 2017 by Teachoo

Last updated at March 8, 2017 by Teachoo

Transcript

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 42+ 143𝑛 is 6 : 1 We know that General term of expansion (a + b)n Tr+1 = nCr an – rbr We need to calculate fifth term from beginning of expansion 42+ 143𝑛 i.e. we need to calculate T5 = T4+1 Putting r = 4 , a = 42 , b = 143 T4+1 = nC442n – 4.1434 T5 = nC4 ((2)14)n – 4 .13144 = nCr (2)14 ×(𝑛−4) .1314 ×4 = nCr (2)14 ×(𝑛−4) .13 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 Now we need to calculate fifth term from end Hence 5th term from end = (n – 3)th term from beginning Another method, rth term from end = (n – r + 2)th term from stating Hence, 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 (42 + 143 )r Tr + 1 = nCr an – rbr Putting r = (n – 3) – 1 = n – 4 a = 42 , b = 143 T(n – 4 + 1) = nCn – 4 42𝑛 −(𝑛−4) . 143𝑛 − 4 = nCn – 4 (2)14n – n + 4 . 1314𝑛 − 4 = nCn – 4 (2)144 . 3−14𝑛 − 4 = nCn – 4 (2)44 .3−(𝑛 − 4)4 = nCn – 4 (2)1 .13𝑛 − 44 Given that ratio fifth term from beginning fifth term from end is 6 : 1 𝐹𝑖𝑓𝑡ℎ 𝑡𝑒𝑟𝑚 𝑓𝑟𝑜𝑚 𝑏𝑒𝑔𝑖𝑛𝑛𝑖𝑛𝑔 𝐹𝑖𝑓𝑡ℎ 𝑡𝑒𝑟𝑚 𝑓𝑟𝑜𝑚 𝑒𝑛𝑑 = 61 𝑇5𝑇𝑛−3 = 61 Putting values from (2) & (3) 𝑛𝐶4 2𝑛−44 . 13𝑛𝐶𝑛−4 2 . 13𝑛−44 = 61 𝑛𝐶4 2𝑛−44 . 13𝑛𝐶𝑛−4 2 . 13𝑛−44 = 61 𝑛𝐶4 𝑛𝐶𝑛−4 . 2𝑛−44 −1 13 1−𝑛−44 = 61 𝑛𝐶4 𝑛𝐶𝑛−4 . 2𝑛−4−44 . 134−𝑛+44 = 61 𝑛𝐶4 𝑛𝐶𝑛−4 . 2𝑛−84 . 138−𝑛4 = 61 𝑛𝐶4 𝑛𝐶𝑛−4 . 2𝑛−84 . 13−(𝑛−8)4 = 61 𝑛𝐶4 𝑛𝐶𝑛−4 . 2𝑛−84. 3𝑛−84 = 61 𝑛𝐶4 𝑛𝐶𝑛−4 . (2 ×3)𝑛−84. = 612 𝑛𝐶4 𝑛𝐶𝑛−4 . 6𝑛−84= 612 Comparing powers of 6 𝑛 − 84 = 12 n – 8 = 42 n – 8 = 2 n = 2 + 8 n = 10 Hence n = 10

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.