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The 6th (from the beginning) term in the expansion of the [2log(103x)+52(x2)log3]m 
is equal to 21. It is known that the binomial coefficient of the 2nd,3rd and 4th term in the expansion are in an A.P., then the value of x is/are 
  1. 0
  2. 1
  3. 2
  4. 3


Solution

The correct options are
A 0
C 2
The coefficients are
mC1,mC2 and mC3
They are T2,T3,T4  terms of an A.P.
2mC2=mC1+mC3m(m1)=m+m(m1)(m2)6m29m+14=0[m0]m=2,7
There should be minimum 6 terms in the expasion, so 
m2,m=7
Now, 
T6=7C5[2log(103x)]75[52(x2)log3]521=212log(103x)2(x2)log31=2log(103x)+log3x20=log(103x)+log3x2(103x)(3x2)=1(103x)3x=932x103x+9=0(3x9)(3x1)=03x=9,1x=2,0

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