Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main20238 Apr 2023Morning ShiftMathematicsBinomial TheoremActual

if the coefficients of three consecutive terms in the expansion of 1 + x n are the ratio 1 : 5 : 20 then the coefficient of the fourth term is

Options

  1. A2436
  2. B5481
  3. C1827
  4. D3654

Correct answer

D. 3654

Step-by-step solution

Given, C r - 1 n : C r n : C r + 1 n = 1 : 5 : 20 Now using the formula C r n C r - 1 n = n - r + 1 r we get, . C r n C r - 1 n = n - r + 1 r = 5 1 ⇒ n - 6 r + 1 = 0     . . . . . . . . . 1 And C r + 1 n C r n = n - r r + 1 = 20 5 ⇒ n - 5 r - 4 = 0     . . . . . . . . . . 2 Solving above equations we get, n = 29   &   r = 5 So, the coefficient of forth term will be C 3 n = C 3 29 = 29 × 28 × 27 3 × 2 × 1 = 3654

Practice Binomial Theorem on Quantrex Academy →

More from Binomial Theorem

If 26 ( 2^3 3 12 2 + 2^5 5 12 4 + 2^7 7 12 6 + + 2¹³ 13 12 12 ) = 3¹³ - , then is equal to: 2026If (1 - x^3)¹⁰ = _ r=0 ¹⁰ a_r x^r (1-x)^ 30-2r , then 9a₉ a₁₀ is equal to __________. 2026If the coefficients of the middle terms in the binomial expansions of (1 + x)²⁶ and (1 - x)²⁸ , 0 , are equal, then the value of is: 2026The coefficient of x^2 in the expansion of (2x^2 + 1 x )¹⁰ , x 0 , is : 2026If the sum of the coefficients of x^7 and x¹⁴ in the expansion of ( 1 x^3 - x^4 )^n , x 0 , is zero, then the value of n is __________. 2026In the expansion of (9x- 1 3 x )¹⁸ , x>0 , if the term independent of x is (221)k , then k is equal to: 2026Let the smallest value of k N , for which the coefficient of x^3 in (1+x)^3 + (1+x)^4 + (1+x)^5 + + (1+x)⁹⁹ + (1+kx)¹⁰⁰ , x 0 , is (43n + 101 4 ) (¹⁰⁰C₃ ) for some n N , be p . The 2026If for 3 r 30 , 30 30-r + 3 30 31-r + 3 30 32-r + 30 33-r = m r , then m equals: 2026 Full Binomial Theorem list All JEE Main PYQs