





Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Examples
Example 2 Important
Example 3 Important
Example 4
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 2 Important Deleted for CBSE Board 2024 Exams
Question 3 Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams You are here
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Last updated at May 29, 2023 by Teachoo
Question 5 The coefficients of three consecutive terms in the expansion of (1 + a)n are in the ratio 1: 7 : 42. Find n. Let the three consecutive terms be (r – 1)th, rth and (r + 1)th terms. i.e. Tr – 1 , Tr & Tr + 1 We know that general term of expansion (a + b)n is Tr + 1 = nCr an – r br For (1 + a)n , Putting a = 1 , b = a Tr+1 = nCr 1n – r ar Tr+1 = nCr ar ∴ Coefficient of (r + 1)th term = nCr For rth term of (1 + a)n Replacing r with r – 1 in (1) Tr – 1 + 1 = nCr – 1 ar – 1 Tr = nCr – 1 ar – 1 ∴ Coefficient of (r)th term = nCr – 1 For (r – 1)th term of (1 + a)n Replacing r with r – 2 in (1) Tr – 2 + 1 = nCr – 2 ar – 2 Tr – 1 = nCr – 2 ar – 2 ∴ Coefficient of (r – 1)th term = nCr – 2 Since the coefficient of (r – 1)th, rth and (r + 1)th terms are in ratio 1 : 7 : 42 (𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 〖(𝒓 − 𝟏)〗^𝒕𝒉 𝒕𝒆𝒓𝒎)/(𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒓^𝒕𝒉 𝒕𝒆𝒓𝒎) = 𝟏/𝟕 〖𝑛𝐶〗_(𝑟 − 2)/〖𝑛𝐶〗_(𝑟 − 1) = 1/7 (𝑛!/((𝑟 − 2)![𝑛 − (𝑟 − 2)]!))/(𝑛!/(𝑟 − 1)!(𝑛 − (𝑟 − 1))!) = 1/7 𝑛!/((𝑟 − 2)![𝑛 − (𝑟 − 2)]!) × ((𝑟 − 1)![𝑛 − (𝑟 − 1)]!)/𝑛! = 1/7 (𝑟 − 1)(𝑟 − 2)!(𝑛 − (𝑟 − 1))!/((𝑟 − 2)! (𝑛 − (𝑟 − 2))!) = 1/7 (𝑟 − 1)(𝑛 − 𝑟 + 1)!/((𝑛 − 𝑟 + 2)!) = 1/7 (𝑟 − 1)(𝑛− 𝑟 + 1)!/((𝑛 − 𝑟 + 2)(𝑛 − 𝑟 + 2 −1)!) = 1/7 (𝑟 − 1)(𝑛− 𝑟 + 1)!/((𝑛 − 𝑟 + 2)(𝑛 − 𝑟 + 1)!) = 1/7 ((𝑟 − 1))/((𝑛 − 𝑟 + 2) ) = 1/7 7(r – 1) = n – r + 2 n – 8r + 9 = 0 Also (𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 𝒓^𝒕𝒉 𝒕𝒆𝒓𝒎)/(𝑪𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒐𝒇 〖(𝒓 + 𝟏)〗^𝒕𝒉 𝒕𝒆𝒓𝒎) = 𝟕/𝟒𝟐 (𝑛!/((𝑟 − 1)![𝑛 − (𝑟 − 1)]!))/(𝑛!/(𝑟! (𝑛 − 𝑟)!)) = 7/42 𝑛!/((𝑟 − 1)!(𝑛 − 𝑟 + 1)!) × (𝑟! (𝑛 − 𝑟)! )/𝑛! = 1/6 (𝑛! × 𝑟 × (𝑟 − 1)!(𝑛 − 𝑟)!)/(𝑛!(𝑟 − 1)! (𝑛 − 𝑟 + 1)!) = 1/6 𝑟(𝑛 − 𝑟)!/((𝑛 − 𝑟 + 1)!) = 1/6 (𝑟 (𝑛 − 𝑟)!)/((𝑛 − 𝑟 + 1) (𝑛 − 𝑟)!) = 1/6 𝑟/(𝑛 + 1 − 𝑟) = 1/6 6r = n + 1 – r n – 7r + 1 = 0 Now we have n – 8r + 9 = 0 …(1) n – 7r + 1 = 0 …(2) From (1) n – 8r + 9 = 0 n = 8r – 9 Putting n = 8r – 9 in (2) (8r – 9) – 7r + 1 = 0 8r – 9 – 7r + 1 = 0 r – 8 = 0 r = 8 Putting value of r in (1) n – 8r + 9 = 0 n – 8(8) + 9 = 0 n – 64 + 9 = 0 n – 55 = 0 n = 55 Hence, n = 55