Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Examples

Example 1

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 You are here

Question 5 Deleted for CBSE Board 2024 Exams

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

Chapter 7 Class 11 Binomial Theorem

Serial order wise

Last updated at May 29, 2023 by Teachoo

Question 4 The second, third and fourth terms in the binomial expansion (x + a)n are 240, 720 and 1080, respectively. Find x, a and n. We know that general term of (a + b)n is Tr+1 = nCr (a)n-r . (b)r Given that second term of (x + a)n is 240 i.e. T2 = 240 T1+1 = 240 Putting r = 1 , a = x & b = a T1+1 = nCr (x)n-1 (a)1 T2 = nCr xn-1a 240 = nCr xn-1a Similarly, Third term of (x + a)n is 720 i.e. T3 = 720 T2+1 = 720 Putting r = 2 , a = x &b = a in (1) T2+1 = nC2 (x)n – 2. (b)2 720 = nC2 xn – 2.a2 Also , Fourth term of (x + a)n is 1080 i.e. T4 = 1080 T3+1 = 1080 Putting r = 3 , a = x & b = a in (1) T3+1 = nC3 xn – 3 . a3 1080 = nC3 xn – 3 . a3 Dividing (3) by (2) 720/240 = (𝑛𝐶2 𝑥^(𝑛 − 2) 𝑎^2)/(𝑛𝐶1 〖𝑥 〗^(𝑛 −1 ) 𝑎) 3 = (𝑛𝐶2 )/(𝑛𝐶1 ) × ( 𝑥^(𝑛−2) )/𝑥^(𝑛−1 ) × ( 𝑎^2)/𝑎 3 = (𝑛!/2!(𝑛−2)!)/(𝑛!/1!(𝑛−1)!) × xn – 2 – (n – 1) × a1 3 = 𝑛!/2!(𝑛−2)! "×" (1!(𝑛−1)! )/𝑛! × xn – 2 – n + 1 × a 3 = 𝑛!/2(𝑛−2)! "×" (1(𝑛−1)(𝑛−2)! )/𝑛! × x–1 × a 3 = ((𝑛−1) )/2 × 𝑎/𝑥 3 × 2 = (n – 1) × 𝑎/𝑥 6 = (n – 1) × 𝑎/𝑥 6/(𝑛 − 1) = 𝑎/𝑥 Now Dividing (4) by (3) 1080/720 = (𝑛𝐶3 𝑥^(𝑛−3) 𝑎^3)/(𝑛𝐶2 〖𝑥^(𝑛−2 ) 𝑎〗^2 ) 3/2 = (𝑛𝐶3 )/(𝑛𝐶2 ) × ( 𝑥^(𝑛−3) )/𝑥^(𝑛−2 ) × 𝑎^3/𝑎^2 3/2 = (𝑛!/3!(𝑛−3)!)/(𝑛!/2!(𝑛−2)!) × xn – 3 – (n – 2) × a3 – 2 3/2 = 𝑛!/3!(𝑛 −3)! × 2!(𝑛−2)!/𝑛! × x-1 × a1 3/2 = 𝑛!/(3(2)!(𝑛 −3)!) × (2!(𝑛−2)(𝑛−3)!)/𝑛! × 𝑎/𝑥 3/2 = ((𝑛 −2))/3 × 𝑎/𝑥 9/(2(𝑛 −2)) = 𝑎/𝑥 Now our equations are 6/(𝑛 − 1) = 𝑎/𝑥 …(A) 9/(2(𝑛 − 2)) = 𝑎/𝑥 …(B) Equating (A) & (B) 9/(2(𝑛 −2)) = 6/(𝑛 −1) 9 × (n – 1) = 6 × 2(n – 2) 9n – 9 = 12n – 24 –9 + 24 = 12n – 9n 15 = 3n 15/3 = n 5 = n n = 5 Putting n = 5 in (A) 6/(𝑛 − 1) = 𝑎/𝑥 6/(5 − 1) = 𝑎/𝑥 6/4 = 𝑎/𝑥 3/2 = 𝑎/𝑥 3x = 2a (3/2)x = a a = (3/2)x Putting n = 5, a = (3/2)x in (2) 240 = nCr xn − 1 a 240 = 5C1 x5 − 1 . (3/2)x 240 = 5!/(1 × 4!) x4 . (3/2)x 240 = 5!/(1 × 4!) x4 . (3/2)x 240 = (5 × 4!)/4! x4 . (3/2)x 240 = 5x4(3/2)x 240 = (5 × 3)/2 × x5 (240 × 2)/(5 × 3) = x5 32 = x5 x5 = 32 x5 = (2)5 x = 2 Putting x = 2 in 𝑎/𝑥 = 3/2 𝑎/2 = 3/2 a = 3 Hence, x = 2, a = 3, n = 5