Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AP EAMCET202418 May 2024Morning ShiftMathematicsBinomial TheoremActual

The square root of independent term in the expansion of ( 2 x^2 5 + 5 x )¹⁰ is

Options

  1. A15 10
  2. B10 15
  3. C30 5
  4. D20 5

Correct answer

C. 30 5

Step-by-step solution

Since the general term in the expansion is ( 2 x^2 5 + 5 x )¹⁰ T _ r+1 = ¹⁰ C _r ( 2 x^2 5 )^ 10-r ( 5^ 1 2 x^ 1 2 )^r= ¹⁰ C _r 2^ 10-r x^ 20-2 r- r 2 5^ 10-r- r 2 for independent of x, 20-2 r- r 2 =0 r=8 So, T ₉= ¹⁰ C ₈ 2^2 5⁻² = 10 9 2 4 5^2=5^3 3^2 2^2 Now, T₉ = 5^3 3^2 2^2 =5 3 2 5 =30 5

Practice Binomial Theorem on Quantrex Academy →

More from Binomial Theorem

If n 13 , n 14 and n 15 are in arithmetic progression, then the positive integer value of ' n ' can be 2026If the coefficients of x^2 and x^3 in the expansion of (3 + kx)^9 are equal, then the value of ' k ' is 2026The remainder when 7¹⁰³ is divided by 25 is 2026_ r=1 ¹⁵ r^2 ( ¹⁵ C_r 15 r-1 )= 20251 81^ n - ^ 2 n C ₁ 10 81^ n + ^ 2 n C ₂ 10^2 81^ n - + 10^ 2 n 81^ n = 2025If x is positive real number and the first negative term in the expansion of (1+ x )^ 27 / 5 is t _ k then k = 2025In the binomial expansion of (p-q)¹⁴ , if the sum of 7^ th term and 8^ th term is zero, then p+q p-q = 2025The numerically greatest term in the expansion of (x+3 y)¹³ , when x= 1 2 and y= 1 3 is 2025 Full Binomial Theorem list All AP EAMCET PYQs