Example 8 - The 2nd, 3rd, 4th terms in expansion (x + a)^n are 240,

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

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.