Q55 / 90 · JEE Main 2019 (09 Apr Shift 2) · Mathematics

If some three consecutive coefficients in the binomial expansion of x+1n in powers of x are in the ratio 2:15:70, then the average of these three coefficients is:

  • A
    227
  • B
    964
  • C
    625
  • Correct answer
    D
    232

Solution

From the given condition

 nCr:nCr+1:nCr+2=2:15:70

⇒ nCr nCr+1=215 and  nCr+1 nCr+2=1570

⇒ n!n-r!·r!n!n-r-1!·r+1!=215 and  n!n-r-1!·r+1!n!n-r-2!·r+2!=1570

⇒n-r-1!·r+1!n-r!·r!=215 and n-r-2!·r+2!n-r-1!·r+1!=314

⇒n-r-1!·r+1·r!n-r·n-r-1!·r!=215 and n-r-2!·r+2·r+1!n-r-1·n-r-2!·r+1!=314

⇒ r+1n-r=215 and  r+2n-r-1=314

⇒ 17r=2n-15 and 17r=3n-31

⇒3n-31=2n-15, ⇒n=16 and r=1

Hence, average = nCr+nCr+1+nCr+23

= 16C1+16C2+16C33

=232.

Hence, the correct answer is option: 'D'.

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